Trigonometry — Grade 11

Q: What is new in Grade 11 Trigonometry?

Grade 11 adds trig identities, reduction formulae, general solutions of trig equations, graphs of the sine, cosine and tangent functions including the effect of the parameters a, k and p, and proving and applying the sine, cosine and area rules for triangles that don't have a right angle.

Q: What is a general solution?

Because sin, cos and tan repeat their values periodically, a trig equation has infinitely many solutions. The general solution captures all of them using a formula with an integer n, such as adding multiples of 360 degrees.

Q: Do I need to know where the sine, cosine and area rules come from, or just how to use them?

Both. CAPS specifically states that the proofs of the sine, cosine and area rules are examinable, not just their application. All three share one construction: dropping an altitude from a vertex splits the triangle into two right triangles you can compare directly.

Q: When do I use the sine rule versus the cosine rule?

Use the sine rule when you know two angles and a side, or two sides and an angle opposite one of them. Use the cosine rule when you know two sides and the included angle, or all three sides.

Q: Do I still need SOH-CAH-TOA and the CAST diagram?

Yes. Grade 11 builds directly on Grade 10's ratio definitions and the CAST diagram; reduction formulae and general solutions both rely on knowing which ratio is positive in which quadrant.

Q: Which resource should I open first for this topic?

Everything you need to learn the topic is on this page already. Once you've been through the notes above, work through the Mastery Bank, then finish with the Test Your Knowledge quiz as a self-check.

Identities, reduction formulae, general solutions of trig equations, graphs of sin/cos/tan and the effect of their parameters, and proving and applying the sine, cosine and area rules for triangles without a right angle. Notes, a mastery bank and a quiz — everything for this topic is one click away.

1. Guided Course 2. Mastery Bank 3. Test Your Knowledge

CAPS alignedGrade 11Paper 2Interactive notes

Grade 11 CAPS Mathematics

Trigonometry

Identities, reduction formulae, general solutions, graphs of sin/cos/tan, and the sine, cosine and area rules.

Your Grade 11 Starting Point

Grade 11 does not restart trigonometry. It extends four Grade 10 ideas into algebra, graphs and non-right triangles.

Read the triangle first

For a terminal arm, label \(x\), \(y\) and \(r\). Then \(\sin\theta=y/r\), \(\cos\theta=x/r\) and \(\tan\theta=y/x\) still explain every sign.

Use an acute reference angle

Special-angle values give the size. The reference angle tells you which familiar value to use before you decide its sign.

Let CAST choose the sign

CAST is a sign check, not a substitute for a diagram. Locate the terminal arm first; then apply the correct positive or negative sign.

Match each side to its angle

In any triangle, side \(a\) is opposite \(\hat A\), side \(b\) is opposite \(\hat B\), and side \(c\) is opposite \(\hat C\). This is essential for the three triangle rules.

The Grade 11 route

You will first turn the Grade 10 ratios into identities. Next, you will use the same coordinate-plane logic for reduction formulae and general solutions. Then you will connect equations to graphs, and finally use an altitude to solve triangles without a right angle.

Trig Identities

Two identities that follow directly from the \((x,y,r)\) definitions.

The two identities

\tan\theta=\frac{\sin\theta}{\cos\theta}\qquad\qquad\sin^2\theta+\cos^2\theta=1

Any point on the terminal arm gives \(\sin\theta=y/r\), \(\cos\theta=x/r\), \(\tan\theta=y/x\) — both identities fall straight out of these three ratios.

Divide the two coordinate ratios

\tan\theta=\frac{y}{x}=\frac{y/r}{x/r}=\frac{\sin\theta}{\cos\theta}

The common radius \(r\) cancels. This is why tangent can be rewritten using sine and cosine.

Divide Pythagoras by \(r^2\)

x^2+y^2=r^2\Rightarrow\left(\frac{x}{r}\right)^2+\left(\frac{y}{r}\right)^2=1

Substitute \(\cos\theta=x/r\) and \(\sin\theta=y/r\) to obtain \(\sin^2\theta+\cos^2\theta=1\).

Domain check

\(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\) only when \(\cos\theta\ne0\), so \(\theta\ne90^\circ+180^\circ n\), \(n\in\mathbb Z\). Before you cancel or divide by a trig expression in a proof, state that it is non-zero.

Choosing an Identity Move

Do not hunt for a formula. Read the expression and make the move that removes the obstacle.

Tangent is present

Rewrite \(\tan\theta\) as \(\sin\theta/\cos\theta\) when this creates a cancellation or puts everything in sin and cos. Keep the original domain restriction.

A squared ratio is in the way

Use \(\sin^2\theta+\cos^2\theta=1\) as \(1-\cos^2\theta=\sin^2\theta\) or \(1-\sin^2\theta=\cos^2\theta\).

A factor might be zero

Factorise first. Do not divide both sides by a trig factor in an equation, because you may throw away a valid solution.

Finish with the original expression

Your final simplified answer must have the same restrictions as the expression you started with, not only the restrictions of the final line.

Try, then check Level 1-4

Simplify: (a) \(\sin\theta\cos\theta\tan\theta\) and (b) \(\dfrac{1-\cos^2\theta}{\sin\theta}\).

Show the identity moves

aReplace tangent: \(\sin\theta\cos\theta\cdot\dfrac{\sin\theta}{\cos\theta}=\boxed{\sin^2\theta}\), provided \(\cos\theta\ne0\).
bUse \(1-\cos^2\theta=\sin^2\theta\): \(\dfrac{\sin^2\theta}{\sin\theta}=\boxed{\sin\theta}\), provided \(\sin\theta\ne0\). The original fraction is undefined at \(\theta=180^\circ n\).
cHarder — simplify \(\dfrac{\cos^2\theta}{1-\sin\theta}\): recognise \(\cos^2\theta=1-\sin^2\theta\), a difference of two squares, so \(\cos^2\theta=(1-\sin\theta)(1+\sin\theta)\). Then \(\dfrac{(1-\sin\theta)(1+\sin\theta)}{1-\sin\theta}=\boxed{1+\sin\theta}\), provided \(\sin\theta\ne1\) (Level 3-4: this needs move B applied inside a factorising step, not on its own).

Quick Check

\(\tan\theta\) is defined as:

Simplify \(\cos^2\theta+\sin^2\theta\)

Why a \(90^\circ\) Shift Switches the Ratio

Build the rule from one right triangle, so you do not have to guess which function changes.

Use the other acute angle

In a right triangle, if one acute angle is \(\theta\), the other one is \(90^\circ-\theta\). The two angles look at the same three sides from different corners.

Notice the side swap

The side opposite \(90^\circ-\theta\) is adjacent to \(\theta\). The side adjacent to \(90^\circ-\theta\) is opposite \(\theta\). The hypotenuse does not change.

Write the co-ratio

\sin(90^\circ-\theta)=\frac{\text{adjacent to }\theta}{\text{hypotenuse}}=\cos\theta

The same swap gives \(\cos(90^\circ-\theta)=\sin\theta\).

Then use CAST for the sign

The ratio name swaps because of the right triangle. A \(+\theta\) or \(-\theta\) in the final angle still determines the sign through the quadrant.

Draw it first

When you see \(90^\circ\pm\theta\), say co-ratio before you simplify: sin and cos swap; tangent becomes its reciprocal. Then locate the final angle to confirm the sign.

Co-ratios: The \(90^\circ\pm\theta\) Family

The swap is only the first decision. The terminal arm still controls the sign.

Apply the co-ratio rule

\sin(90^\circ-\theta)=\cos\theta\qquad \cos(90^\circ-\theta)=\sin\theta\]\[\sin(90^\circ+\theta)=\cos\theta\qquad \cos(90^\circ+\theta)=-\sin\theta\]\[\tan(90^\circ-\theta)=\frac1{\tan\theta}\qquad \tan(90^\circ+\theta)=-\frac1{\tan\theta}

For tangent co-ratios, both sides must be defined. Check the original expression before simplifying.

Angle family	What happens to the ratio name?	What decides the sign?
\(90^\circ\pm\theta\)	It switches: sin \(\leftrightarrow\) cos; tan becomes its reciprocal.	The quadrant of the final angle.
\(180^\circ\pm\theta,\ 360^\circ\pm\theta,\ -\theta\)	It stays the same.	CAST gives the sign.

Watch out

\(\sin(180^\circ-\theta)\ne\cos\theta\). The ratio only switches in the \(90^\circ\) family. For \(180^\circ\), \(360^\circ\) and negative-angle reductions, keep the same ratio name and use CAST for the sign.

Worked example Level 2

Simplify \(\cos(90^\circ+\theta)+\sin(90^\circ-\theta)\).

Show solution

1\(\cos(90^\circ+\theta)=-\sin\theta\).
2\(\sin(90^\circ-\theta)=\cos\theta\).
3\(\boxed{\cos\theta-\sin\theta}\).

Reduction Formulae

CAST as a mirror — every angle reduces to a reference angle \(\theta\).

Reading the diagram: the angle \(180°-\theta\) lands in Quadrant 2, with reference angle \(\theta\) measured back to the negative x-axis. In Q2, only sin is positive — so \(\sin(180°-\theta)=\sin\theta\) but \(\cos(180°-\theta)=-\cos\theta\).

	\(180°-\theta\)	\(180°+\theta\)	\(360°-\theta\)	\(360°+\theta\)	\(-\theta\)
\(\sin\)	\(\sin\theta\)	\(-\sin\theta\)	\(-\sin\theta\)	\(\sin\theta\)	\(-\sin\theta\)
\(\cos\)	\(-\cos\theta\)	\(-\cos\theta\)	\(\cos\theta\)	\(\cos\theta\)	\(\cos\theta\)
\(\tan\)	\(-\tan\theta\)	\(\tan\theta\)	\(-\tan\theta\)	\(\tan\theta\)	\(-\tan\theta\)

The method, every time

1. Find which quadrant the angle lands in. 2. Find the reference angle (the acute angle back to the x-axis). 3. Use CAST to decide the sign. In this \(180^\circ\)/\(360^\circ\)/negative-angle family, the ratio name stays the same; the \(90^\circ\pm\theta\) co-ratio family is on the previous slide.

Worked example Level 2-3

Simplify: \(\sin(180°-\theta)+\cos(360°-\theta)\)

Show solution

1\(\sin(180°-\theta)=\sin\theta\) (Q2, sin positive)
2\(\cos(360°-\theta)=\cos\theta\) (Q4, cos positive)
3Sum: \(\boxed{\sin\theta+\cos\theta}\)

Reduction Formulae: A Reliable Decision Routine

Use the same four checks for every reduction question. This prevents a correct sign with the wrong ratio name.

Spot the angle family

Is the angle built around \(90^\circ\)? If yes, it is a co-ratio: sin and cos swap. If not, keep the original ratio name.

Locate the terminal arm

Use the \(180^\circ\), \(360^\circ\) or negative-angle part to identify the quadrant of the final angle.

Use the reference angle

The small acute angle back to the x-axis is \(\theta\). It gives the familiar ratio value or expression you will write.

Apply CAST last

Give the reduced ratio the sign of the final quadrant. The sign belongs to the whole term, including any subtraction outside brackets.

One reduction at a time Level 1

Using the four-step routine, simplify \(\sin(180^\circ+\theta)\).

Show solution

1Not built around \(90^\circ\), so the ratio name stays sine.
2\(180^\circ+\theta\) lands in Quadrant 3.
3Reference angle: \(\theta\).
4In Q3, sine is negative: \(\boxed{\sin(180^\circ+\theta)=-\sin\theta}\)

Mixed family example Level 3

Simplify \(\sin(180^\circ-\theta)-\cos(90^\circ+\theta)+\tan(360^\circ+\theta)\).

Keep sine

\(\sin(180^\circ-\theta)=\sin\theta\): Q2 makes sine positive.

Swap cosine

\(\cos(90^\circ+\theta)=-\sin\theta\): the \(90^\circ\) family swaps the name and Q2 makes cosine negative.

Keep tangent

\(\tan(360^\circ+\theta)=\tan\theta\): one full turn returns to the same quadrant.

Respect the outside sign

\(\sin\theta-(-\sin\theta)+\tan\theta=\boxed{2\sin\theta+\tan\theta}\).

Quick Check

\(\cos(180°+\theta)\) simplifies to:

\(\sin(-\theta)\) simplifies to:

\(\cos(90^\circ+\theta)\) simplifies to:

How to Build a General Solution

A general solution is not a formula to copy. It is a set of correct angles repeated by the function's period.

Isolate one trig ratio

Get the equation into \(\sin\theta=k\), \(\cos\theta=k\), or \(\tan\theta=k\). Check that \(-1\leq k\leq1\) for sine or cosine.

Find the acute reference angle

Use the positive size of \(k\): \(\alpha=\sin^{-1}|k|\), \(\cos^{-1}|k|\), or \(\tan^{-1}|k|\). The reference angle must be acute.

Choose quadrants from the sign

Use CAST. A positive or negative value decides the two possible quadrants for sine/cosine, or the one repeating tangent family.

Add the correct period

Sine and cosine repeat after \(360^\circ\). Tangent repeats after \(180^\circ\). Write \(n\in\mathbb Z\) for all whole-number repeats.

Ratio	Positive value	Negative value	Repeat
Sine	\(\alpha,\ 180^\circ-\alpha\)	\(180^\circ+\alpha,\ 360^\circ-\alpha\)	Add \(360^\circ n\) to each family.
Cosine	\(\alpha,\ 360^\circ-\alpha\)	\(180^\circ-\alpha,\ 180^\circ+\alpha\)	Add \(360^\circ n\) to each family.
Tangent	\(\theta=\alpha+180^\circ n\)	\(\theta=(180^\circ-\alpha)+180^\circ n\)	One family is enough because tangent repeats after \(180^\circ\).

Do not lose signs

The table is a check, not a replacement for CAST. For a negative sine value, for example, the angles must land below the x-axis in Quadrants 3 and 4; \(\alpha\) and \(180^\circ-\alpha\) would be wrong.

General Solutions: See the Two Families

The two points on the circle show why a positive sine value has two answers in one full turn.

Both points sit at the same height (\(y=0{,}5\)) — the reference angle \(30°\) applies in both Q1 and Q2, since sin is positive in each.

Worked example Level 3

Solve generally: \(\sin\theta=0{,}5\)

Show solution

1Reference angle: \(\sin^{-1}(0{,}5)=30°\)
2\(\boxed{\theta=30°+n\cdot360°\text{ or }\theta=150°+n\cdot360°,\;n\in\mathbb{Z}}\)
3Check: \(\sin150°=\sin(180°-30°)=\sin30°=0{,}5\) ✓

Worked example: a specific interval Level 3

Solve \(2\cos\theta-\sqrt3=0\) for \(\theta\in[-180^\circ;360^\circ]\).

Show solution

1\(2\cos\theta=\sqrt3\Rightarrow\cos\theta=\dfrac{\sqrt3}{2}\). The reference angle is \(30^\circ\).
2Cosine is positive in Quadrants 1 and 4: \(\theta=30^\circ+360^\circ n\) or \(\theta=-30^\circ+360^\circ n\).
3Keep only values in the requested interval: \(\boxed{\theta\in\{-30^\circ;30^\circ;330^\circ\}}\).

Quick Check

The general solution of \(\tan\theta=k\) uses steps of:

If \(\cos\theta=-0{,}5\), the reference angle is \(60°\). One family of the general solution is:

Graphs of the Trig Functions

Sine, cosine and tangent, each with their own shape and period.

\(y=\sin x\): period \(360°\), amplitude \(1\), range \([-1,1]\), passes through the origin.

\(y=\cos x\)

Same period (\(360°\)) and amplitude (\(1\)) as sine, but shifted — starts at its maximum \((0°,1)\) instead of at zero.

\(y=\tan x\)

Period \(180°\) (half of sine/cosine), passes through the origin, and has asymptotes at \(90°\) and \(270°\) where the function is undefined.

Required plotting interval

You must be able to plot the three base graphs on \([-360^\circ;360^\circ]\), not only on one cycle. Use the period to repeat the same key points to the left and right of the origin; tangent must stay in separate branches between every asymptote.

Sketch a Trig Graph From Key Points

Do not freehand from memory. Place the structural points first, then draw the curve through them.

Base graph	One cycle: key x-values	What to label
\(y=\sin x\)	\((0^\circ,0),(90^\circ,1),(180^\circ,0),(270^\circ,-1),(360^\circ,0)\)	Midline \(y=0\), maximum \(1\), minimum \(-1\), period \(360^\circ\).
\(y=\cos x\)	\((0^\circ,1),(90^\circ,0),(180^\circ,-1),(270^\circ,0),(360^\circ,1)\)	Midline \(y=0\), maximum \(1\), minimum \(-1\), period \(360^\circ\).
\(y=\tan x\)	\((0^\circ,0)\), repeat at \(180^\circ\); branches approach \(90^\circ\) and \(270^\circ\).	Vertical asymptotes at \(90^\circ+180^\circ n\); no maximum or minimum; period \(180^\circ\).

Set up the axes

Use the required interval, usually \([-360^\circ;360^\circ]\), and label the important angle values before plotting points.

Find the period and amplitude

For \(\sin\) and \(\cos\), period \(=360^\circ/k\). For \(\tan\), period \(=180^\circ/k\). The amplitude of sine/cosine is \(|a|\).

Move every key point

Apply the horizontal shift before you plot: \(y=a\sin(x+p)\) moves left by \(p\) when \(p>0\), and right when \(p<0\).

Draw the correct shape

Join sine and cosine points smoothly. For tangent, draw separate increasing branches and never let the curve cross an asymptote.

The two crossings visible here are exactly the two solutions found earlier by general solution — the graph and the algebra agree.

Reading the basics off the graph Level 1

Use the graph of \(y=\sin x\) above to state its amplitude and period.

Show solution

1The curve reaches a maximum of \(1\) and a minimum of \(-1\), so the amplitude is \(\boxed{1}\).
2One full cycle finishes at \(x=360°\), so the period is \(\boxed{360°}\).

Reading the graph Level 2-3

Use the graph of \(y=\sin x\) above to confirm the two solutions of \(\sin x=0{,}5\) found earlier by general solution.

Show solution

1Draw a horizontal line at \(y=0{,}5\). It crosses the curve at two points within one cycle.
2Reading off the graph, the crossings are at \(x=30°\) and \(x=150°\) — matching the general solution exactly.
3The graph makes it visually obvious why there are always two solutions per cycle for sine: the curve is symmetric about \(x=90°\).

The Effect of \(a\), \(k\) and \(p\)

Stretch, squeeze and slide the base graphs.

Parameter \(k\) changes the period: \(y=\sin(kx)\) completes \(k\) full cycles in the space one used to take — here \(k=2\) halves the period from \(360°\) to \(180°\). The general rule: period \(=\dfrac{360°}{k}\) for sin/cos, \(\dfrac{180°}{k}\) for tan.

Effect of \(a\)

Stretches (\(|a|>1\)) or compresses (\(0<|a|<1\)) the graph vertically — the amplitude becomes \(|a|\). If \(a<0\), the graph also flips upside down.

Effect of \(p\)

Shifts the whole graph horizontally: \(y=\sin(x+p)\) moves the base graph \(p°\) to the left (same \(x+p\) sign convention as parabola/hyperbola/exponential shifts in Functions & Graphs — watch the sign).

Start with the parent graph

Name the base function first: \(\sin\), \(\cos\) or \(\tan\). Its key points, asymptotes and starting position do not disappear.

Read \(a\) vertically

Amplitude is \(|a|\) for sine and cosine. A negative \(a\) reflects the graph in the x-axis; it does not change the period.

Read \(k\) horizontally

Calculate the new period before plotting: \(360^\circ/k\) for sine/cosine, \(180^\circ/k\) for tangent.

Read \(p\) from the bracket

Set the inside bracket equal to zero. For \(x+p=0\), the graph's starting feature occurs at \(x=-p\), which avoids guessing the shift direction.

Sketch order

Find the period and the horizontal shift first; move the key x-values; then apply the vertical amplitude or reflection. Grade 11 sketch questions change at most two parameters at a time, so make one deliberate change at a time.

\(a=2\) doubles the amplitude; \(p=-30°\) shifts the curve \(30°\) to the right (since \(x+p=x-30°\) means \(p=-30°\)).

Worked example — one parameter only Level 1

Describe the transformation from \(y=\sin x\) to \(y=3\sin x\), and state the new amplitude.

Show solution

1Compare to \(y=a\sin x\): here \(a=3\), and there is no horizontal shift.
2\(a=3\) stretches the graph vertically: \(\boxed{\text{new amplitude}=3}\) (range becomes \([-3,3]\)).

Worked example — two parameters together Level 3

Describe the transformation from \(y=\sin x\) to \(y=2\sin(x-30°)\), and state the new amplitude and where the graph now crosses its midline going upward.

Show solution

1Compare to the general form \(y=a\sin(x+p)\): here \(a=2\) and \(x+p=x-30°\Rightarrow p=-30°\).
2\(a=2\) doubles the amplitude: new amplitude \(=2\) (range becomes \([-2,2]\)).
3\(p=-30°\) shifts every feature of the base graph \(30°\) to the right (the sign is opposite to \(p\)'s own sign, exactly like the parabola/hyperbola \(x+p\) convention).
4\(y=\sin x\) crosses its midline going upward at \(x=0°\), so the transformed graph does the same at \(\boxed{x=30°}\).

Quick Check

What is the period of \(y=\cos(3x)\)?

\(y=\sin(x+45°)\) is the base sine graph shifted:

Quick Check

The period of \(y=\tan x\) is:

At \(x=0°\), the graph of \(y=\cos x\) passes through:

Beyond Right Triangles

SOH-CAH-TOA only works when there's a right angle.

Naming convention: each lowercase side letter is opposite the matching uppercase vertex — side \(a\) is opposite \(\hat A\), and so on.

Facts you can identify	Your first move
A right angle is marked	Use SOH-CAH-TOA or Pythagoras first. Do not reach for a non-right-triangle rule automatically.
A complete opposite pair, or two angles and one side	Use the sine rule. If needed, find the third angle first so you can match a side to its opposite angle.
Two sides and the included angle (SAS), or all three sides (SSS)	Use the cosine rule to find a missing side or angle.
Two sides and their included angle, and the question asks for area	Use the area rule: \(\tfrac12ab\sin C\).

Label before choosing

First match every side to the angle opposite it. Then list only the facts you know. A formula is chosen by the pattern of information, not because its name looks familiar.

Proving the Rules

One construction — drop an altitude — unlocks all three proofs. You need to understand where each formula comes from, not only substitute into it.

Dropping the altitude \(h\) from \(C\) splits \(\triangle ABC\) into two right triangles, \(ACD\) and \(BCD\), that share the same height \(h\) — every rule below falls out of comparing them.

What the construction gives you immediately

h=b\sin A=a\sin B \qquad\qquad AD=b\cos A

The first comes from right triangle \(ACD\) (angle \(A\), hypotenuse \(b\)) equalling right triangle \(BCD\) (angle \(B\), hypotenuse \(a\)) — both express the same height \(h\). The proofs on the next three slides all start here.

Draw the altitude

Name the foot \(D\) and the height \(h\). One non-right triangle has now become two right triangles.

Write what each small triangle says

Use SOH-CAH-TOA in both small triangles. They must describe the same \(h\), so their expressions can be equated.

Choose the algebra move

Equate two height expressions for the sine rule; combine Pythagoras with \(AD=b\cos A\) for the cosine rule; use \(\tfrac12\text{base}\times\text{height}\) for area.

Finish in the correct form

State the formula using the matching opposite side-angle labels. Do not skip from the sketch to the final formula without the connecting algebra.

The Sine Rule

Use when you know two angles and a side, or two sides and a non-included angle.

Drawn to scale from the given values — the longer side sits opposite the larger angle, exactly as the sine rule predicts.

Read the facts

You know \(A=40°\), \(B=65°\), and side \(a=10\), which is opposite \(A\). That complete opposite pair tells you to use the sine rule to find \(b\), opposite \(B\).

Formula

\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}

Proof chain Level 4

1Drop the altitude \(h\) from \(C\) to \(D\) on \(AB\).
2In right triangle \(ACD\), \(h=b\sin A\). In right triangle \(BCD\), \(h=a\sin B\).
3Both expressions are the same height: \(b\sin A=a\sin B\). Rearranging gives \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}\).
4A second altitude gives \(\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\), so \(\boxed{\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}}\).

SSA check

When you use the sine rule with a known angle-side pair and another side (SSA), \(\sin^{-1}\) can give an angle \(B\) or \(180^\circ-B\). Test both; keep a second triangle only if the remaining angle is positive and the diagram is possible.

Worked example Level 3

In \(\triangle ABC\), \(\hat A=40°\), \(\hat B=65°\), \(a=10\). Determine \(b\).

Show solution

1\(\dfrac{b}{\sin B}=\dfrac{a}{\sin A} \Rightarrow b=\dfrac{a\sin B}{\sin A}\)
2\(b=\dfrac{10\sin65°}{\sin40°}\approx\dfrac{10(0{,}9063)}{0{,}6428}\approx\boxed{14{,}10}\)

The Cosine Rule

Use when you know two sides and the included angle, or all three sides.

Two sides and the included angle are given (SAS) — exactly the pattern the cosine rule is built for.

Read the facts

The known sides \(b=8\) and \(c=6\) meet at the included angle \(A=60°\). There is no known opposite pair, so this SAS pattern calls for the cosine rule.

Formula

a^2=b^2+c^2-2bc\cos A

For an angle from three known sides, rearrange first: \(\cos A=\dfrac{b^2+c^2-a^2}{2bc}\), then use \(A=\cos^{-1}(\text{value})\).

Proof chain Level 4

1Use the altitude construction: \(AD=b\cos A\), so \(BD=c-b\cos A\).
2In right triangle \(BCD\), Pythagoras gives \(a^2=BD^2+h^2=(c-b\cos A)^2+(b\sin A)^2\).
3Expand: \(a^2=c^2-2bc\cos A+b^2\cos^2A+b^2\sin^2A\).
4Use \(\cos^2A+\sin^2A=1\): \(\boxed{a^2=b^2+c^2-2bc\cos A}\).

Worked example Level 3

In \(\triangle ABC\), \(b=8\), \(c=6\), \(\hat A=60°\). Determine \(a\).

Show solution

1\(a^2=8^2+6^2-2(8)(6)\cos60°=64+36-96(0{,}5)\)
2\(a^2=100-48=52\)
3\(a=\sqrt{52}=2\sqrt{13}\approx\boxed{7{,}21}\)

The Area Rule

Use when you know two sides and the included angle.

The same "two sides + included angle" information as the cosine rule — but here we want the area, not a missing side.

Read the facts

You know sides \(a=7\) and \(b=9\) and their included angle \(C=50°\). Because the question asks for area, select the area rule rather than the cosine rule.

Formula

\text{Area}=\frac12bc\sin A=\frac12ac\sin B=\frac12ab\sin C

Choose the version that uses the two known sides and their included angle.

Proof chain Level 4

1Area of any triangle is \(\dfrac12\times\text{base}\times\text{height}\). Take \(AB=c\) as the base.
2The altitude gives \(h=b\sin A\).
3Substitute the height: \(\text{Area}=\dfrac12\times c\times b\sin A=\dfrac12bc\sin A\).
4Choose a different base to obtain \(\boxed{\text{Area}=\dfrac12bc\sin A=\dfrac12ac\sin B=\dfrac12ab\sin C}\).

Worked example Level 2-3

In \(\triangle ABC\), \(a=7\), \(b=9\), \(\hat C=50°\). Determine the area.

Show solution

1Area \(=\dfrac12(7)(9)\sin50°=31{,}5\times0{,}7660\)
2\(\approx\boxed{24{,}13\text{ square units}}\)

Worked example — working backward from the area Level 4

In \(\triangle ABC\), the area is \(15\text{ cm}^2\), \(a=6\text{ cm}\) and \(b=8\text{ cm}\). Determine the size of the acute angle \(\hat C\).

Show solution

1This time the area is given and the angle is unknown — substitute into the same formula, but now solve for the angle: \(15=\dfrac12(6)(8)\sin C\)
2\(15=24\sin C\Rightarrow\sin C=\dfrac{15}{24}=0{,}625\)
3\(\boxed{\hat C\approx38{,}68°}\) (taking the acute angle, as asked)

Quick Check

You know all three sides of a triangle but no angles. Which rule finds an angle?

In a triangle with \(a=5\), \(b=6\), included angle \(C=40°\), the area is:

Putting It Together

Every Grade 11 trig question is one of these decisions.

Ask yourself	Then
Need to simplify an expression?	Try the identities \(\tan\theta=\sin\theta/\cos\theta\) and \(\sin^2\theta+\cos^2\theta=1\) first.
Angle is \(180°\pm\theta\), \(360°\pm\theta\) or \(-\theta\)?	Find the quadrant, use CAST for the sign, keep the same function.
Solving \(\sin\theta=k\) / \(\cos\theta=k\) / \(\tan\theta=k\)?	Find the reference angle, then write the correct general solution family.
Need to sketch or read a trig graph?	Know the period, amplitude and starting value for sin, cos and tan, and how \(a\) (amplitude), \(k\) (period) and \(p\) (shift) each change them.
Triangle has no right angle?	Sine rule (angle+side pairs), cosine rule (SAS/SSS), or area rule (SAS for area) — and know each one's altitude-drop proof, not just how to apply it.

Secure the easy marks

Know exact values, basic identities, parent-graph facts and the three formulae without hesitation.

Carry out a routine

Reduce an angle, solve a familiar equation, calculate a side or plot known key points accurately.

Make the method choice

Choose the rule from the diagram, apply CAST in a mixed expression, filter an interval, or read a transformed graph.

Explain and connect

Write a triangle-rule proof, justify a restriction, or connect algebra, a graph and a diagram in one multi-step question.

Looking ahead

Grade 12 adds the compound angle and double angle identities, general proving of identities, and extends problem-solving from 2D into 3D.

1 / 27

What CAPS Expects You to Know

The Grade 11 Trigonometry knowledge statement this page is built from.

1
Derive and use \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\), including its domain, and \(\sin^2\theta+\cos^2\theta=1\).
2
Derive and use the reduction formulae for \(90^\circ\pm\theta\), \(180^\circ\pm\theta\), \(360^\circ\pm\theta\) and \(-\theta\).
3
Determine where an identity is defined, then write general solutions and solutions in a specified interval.
4
Plot, sketch and interpret trig graphs, including the effects of \(a\), \(k\) and \(p\), with at most two parameters changing at once.
5
Prove and apply the sine, cosine and area rules.
6
Solve problems in two dimensions using those three rules.

How to Use This Lesson

A few practical notes before you start.

Make sure Grade 10's CAST diagram is solid first — reduction formulae and general solutions both depend on it.
For reduction formulae, always find the quadrant and reference angle before deciding the sign.
Sketch a quick trig graph whenever you're unsure how many solutions an equation should have.
The sine, cosine and area rule proofs share one trick — drop an altitude — learn that once and all three follow.
Try each worked example yourself before pressing “Show solution.”

Learn It in Short Videos

Four free, independent videos — not made by Equation Station SA.

Trig Identities

Using the Pythagorean identity and the tan/sin/cos relationship.

Khan Academy · Pythagorean trig identity from soh cah toa

Reduction Formulae

Using rotations and symmetry to relate angles in different quadrants.

Khan Academy · Relating trig function through angle rotations

Trig Graphs

Sketching and interpreting the sine, cosine and tangent curves.

Khan Academy · Midline, amplitude and period of a function

Sine & Cosine Rules

Solving non-right triangles with the sine rule and cosine rule.

Khan Academy · Law of sines

Want another explanation? Find a lesson or activity for your subtopic below.

Learn More by Subtopic

Use a second explanation when you get stuck, then try the same skill without help.

These links open other websites in a new tab. Siyavula follows the South African grade sequence. International resources may also use radians; use degrees for the questions on this page.

Refresh Your Triangle Ratios

Revisit side labels and inverse trig before moving on to identities and equations.

Khan Academy: right-triangle revision and practice

Pythagorean & Quotient Identities

Work through a proof one side at a time. Write the identity used next to each change.

Siyavula: trig identities and worked proofs

Reduction Formulae, Co-ratios & CAST

For each expression, name the quadrant, decide the sign, then decide whether the ratio changes.

General Solutions & Given Intervals

Write every solution family before selecting the values that belong to the stated interval. Check both endpoints.

Siyavula: trigonometric equations and general solutions

Sine, Cosine & Area Rules

Compare a known opposite pair, two sides with their included angle, and three known sides. Choose a rule before calculating.

Graphs, Period & Transformations

In Desmos, choose Degrees in Graph Settings and show 0 to 360 on the x-axis. Compare y=sin(x) with y=2sin(2x). Predict the amplitude and period first.

Ready to practise? Return to the Grade 11 Mastery Bank.

Common Exam Mistakes

Avoid these errors. They cost marks every year.

Forgetting the sign in a reduction formula

Always check CAST for the quadrant before writing the answer — don't just guess.

Only writing one family of a general solution

Both sin and cos equations need two families of solutions, not one.

Using the sine rule when the cosine rule is needed

Sine rule needs an angle-side pair; if you only have three sides or two sides + the included angle, use the cosine rule instead.

Mixing up amplitude and period

Amplitude (parameter \(a\)) is how tall the graph is; period (parameter \(k\)) is how often it repeats — they change independently.

Only being able to apply a rule, not prove it

The sine, cosine and area rule proofs are explicitly examinable in Grade 11 — know the altitude construction, not just the formulas.

Practise This Topic

You've done the notes above — now practise and test yourself.

BANK

Mastery Bank

Grade 11 Trigonometry Mastery Bank

Exam-style Grade 11 questions arranged by level, combining original practice with clearly identified paper-and-memo matches.

Step 1 • Practise • Cognitive Levels

Open Mastery Bank

TEST

Test Your Knowledge

Trigonometry Grade 11 Test Your Knowledge

Auto-marked quiz with instant feedback, explanations and a complete answer review.

Step 2 • Test Yourself • Auto-Marked + Review

Test Your Knowledge

Student Questions

Frequently Asked Questions

Straight answers to common Grade 11 CAPS questions about trigonometry.

What is new in Grade 11 Trigonometry?

Grade 11 adds trig identities, reduction formulae, general solutions of trig equations, graphs of the sine, cosine and tangent functions including the effect of the parameters a, k and p, and proving and applying the sine, cosine and area rules for triangles that don't have a right angle.

What is a general solution?

Because sin, cos and tan repeat their values periodically, a trig equation has infinitely many solutions. The general solution captures all of them using a formula with an integer n, such as adding multiples of 360 degrees.

Do I need to know where the sine, cosine and area rules come from, or just how to use them?

Both. CAPS specifically states that the proofs of the sine, cosine and area rules are examinable, not just their application. All three share one construction: dropping an altitude from a vertex splits the triangle into two right triangles you can compare directly.

When do I use the sine rule versus the cosine rule?

Use the sine rule when you know two angles and a side, or two sides and an angle opposite one of them. Use the cosine rule when you know two sides and the included angle, or all three sides.

Do I still need SOH-CAH-TOA and the CAST diagram?

Yes. Grade 11 builds directly on Grade 10's ratio definitions and the CAST diagram; reduction formulae and general solutions both rely on knowing which ratio is positive in which quadrant.

Which resource should I open first for this topic?

Everything you need to learn the topic is on this page already. Once you've been through the notes above, work through the Mastery Bank, then finish with the Test Your Knowledge quiz as a self-check.