Trig Identities & R-Formula | A Maths Study Guide

Secondary 3 & 4 Additional Mathematics · Study guide

Build identities from the unit circle, preserve their domains and use R-formula to combine sine and cosine.

Concept → worked examples → practice28 min read · practice linked below

An identity is true wherever both sides are defined

This guide takes you from Secondary 3 basic identities to Secondary 4 addition, double-angle and R-formula methods. An equation is solved for particular angles. An identity is a relationship that holds for every angle in its stated domain. In a proof, begin with one side and transform it into the other using valid steps.

On the unit circle, the point at angle $\theta$ is $(\cos\theta,\sin\theta)$ . Its distance from the origin is 1, so Pythagoras gives $\sin^2\theta+\cos^2\theta=1$ for every real angle. Dividing by a nonzero square gives the other Pythagorean identities.

The right triangle illustrates an acute angle. Signed unit-circle coordinates extend the identity to every real angle.

The six ratios and their restrictions
Identity or definition	Condition
$\tan\theta=\frac{\sin\theta}{\cos\theta}$ , $\sec\theta=\frac1{\cos\theta}$	$\cos\theta\ne0$
$\cot\theta=\frac{\cos\theta}{\sin\theta}$ , $\operatorname{cosec}\theta=\frac1{\sin\theta}$	$\sin\theta\ne0$
$1+\tan^2\theta=\sec^2\theta$	$\cos\theta\ne0$
$1+\cot^2\theta=\operatorname{cosec}^2\theta$	$\sin\theta\ne0$

The reciprocal form $\cot\theta=1/\tan\theta$ is useful only where that right-hand expression is defined. For example, $\cot90^\circ=0$ from $\cos\theta/\sin\theta$ , even though $\tan90^\circ$ is undefined. Cancelling a factor never restores an angle excluded from the original expression.

Example 1: simplify and preserve the domain

Simplify $(1-\cos^2\theta)/\sin\theta$ .

The original denominator requires $\sin\theta\ne0$ . Using $1-\cos^2\theta=\sin^2\theta$ ,

\frac{1-\cos^2\theta}{\sin\theta}=\frac{\sin^2\theta}{\sin\theta}=\sin\theta,\qquad \sin\theta\ne0.

Although $\sin\theta$ itself is defined at $\theta=0$ , the original fraction is not. That exclusion remains part of the answer.

Example 2: prove an identity from one side

Prove $\sec\theta-\cos\theta=\sin\theta\tan\theta$ , where $\cos\theta\ne0$ .

\begin{aligned}\text{LHS}&=\frac1{\cos\theta}-\cos\theta\\&=\frac{1-\cos^2\theta}{\cos\theta}\\&=\frac{\sin^2\theta}{\cos\theta}\\&=\sin\theta\tan\theta=\text{RHS}.\end{aligned}

Each equality is valid on the given domain. Converting to sine and cosine exposed a common denominator and then the Pythagorean identity. Checking a few angles can detect a mistake, but it does not prove an identity.

Addition and subtraction formulae

Adding angles rotates the unit-circle point. Resolving the rotated point into horizontal and vertical components gives

\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B,

\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B.

For cosine, the sign between the products is the opposite of the sign inside the angle. Dividing the sine expansion by the cosine expansion gives

\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B},

provided $\tan A$ , $\tan B$ and the displayed quotient are defined. If a tangent expression is undefined at an intermediate angle, the sine and cosine formulae may still be used. All angles in a calculation must use the same unit.

Example 3: exact values without a calculator

Find exact values of $\sin15^\circ$ and $\tan75^\circ$ .

\sin(45^\circ-30^\circ)=\frac{\sqrt2}{2}\frac{\sqrt3}{2}-\frac{\sqrt2}{2}\frac12=\frac{\sqrt6-\sqrt2}{4}.

\tan(45^\circ+30^\circ)=\frac{1+1/\sqrt3}{1-1/\sqrt3}=\frac{\sqrt3+1}{\sqrt3-1}=2+\sqrt3.

The tangent denominator is nonzero. Both answers are positive, as expected for acute angles; $\tan75^\circ>1$ because 75° is above 45°.

Example 4: use quadrants before expanding

Given $\sin A=3/5$ , $90^\circ<A<180^\circ$ , $\cos B=12/13$ and $0^\circ<B<90^\circ$ , find $\sin(A+B)$ and $\cos(A+B)$ .

A is in quadrant II, so $\cos A=-\sqrt{1-9/25}=-4/5$ . B is acute, so $\sin B=\sqrt{1-144/169}=5/13$ . Therefore

\sin(A+B)=\frac35\frac{12}{13}+\left(-\frac45\right)\frac5{13}=\frac{16}{65},

\cos(A+B)=\left(-\frac45\right)\frac{12}{13}-\frac35\frac5{13}=-\frac{63}{65}.

Check: $(16/65)^2+(-63/65)^2=1$ . The positive sine and negative cosine place $A+B$ in quadrant II, consistent with these particular angles. Taking both square roots positive would give wrong answers.

Double angles: choose a form that reduces the work

Set $B=A$ in the addition formulae:

\sin2A=2\sin A\cos A,

\cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A,

\tan2A=\frac{2\tan A}{1-\tan^2A}.

The tangent form needs $\cos A\ne0$ and $1-\tan^2A\ne0$ . Choose $1-2\sin^2A$ if you know sine, or $2\cos^2A-1$ if you know cosine. Conversely, $\sin^2A=(1-\cos2A)/2$ and $\cos^2A=(1+\cos2A)/2$ . A square root of these expressions still needs a quadrant to choose its sign.

Example 5: signs of double-angle values

For $\sin A=3/5$ and $90^\circ<A<180^\circ$ , find $\sin2A,\cos2A,\tan2A$ .

As above, $\cos A=-4/5$ . Then

\sin2A=2\left(\frac35\right)\left(-\frac45\right)=-\frac{24}{25},\quad \cos2A=1-2\left(\frac35\right)^2=\frac7{25}.

Since $\cos2A\ne0$ , $\tan2A=(-24/25)/(7/25)=-24/7$ . Check: $24^2+7^2=25^2$ . The angle 2A is in quadrant IV for this value of A; doubling an obtuse angle does not automatically make every ratio negative.

Example 6: a proof using double angles

Prove $\frac{\sin2x}{1+\cos2x}=\tan x$ on its domain.

Since $1+\cos2x=2\cos^2x$ , the original denominator is nonzero precisely when $\cos x\ne0$ . On this domain,

\frac{\sin2x}{1+\cos2x}=\frac{2\sin x\cos x}{2\cos^2x}=\frac{\sin x}{\cos x}=\tan x.

For degree angles, exclude $x=90^\circ+180^\circ n$ , where n is an integer; in radians, exclude $x=\pi/2+n\pi$ . These are domain restrictions, not extra solutions to find.

R-formula: combine two waves of the same frequency

To write $a\sin\theta+b\cos\theta=R\sin(\theta+\alpha)$ , expand the right side and match coefficients:

R\cos\alpha=a,\qquad R\sin\alpha=b,\qquad R=\sqrt{a^2+b^2}.

When a and b are not both zero, choose $R>0$ . The signs of a and b determine the quadrant of $\alpha$ . Using only $\tan\alpha=b/a$ can select the wrong quadrant. For positive a and b, $\alpha$ is acute. A minus sign in the original expression may be handled by a subtraction form, then checked by expanding.

You can equally use $R\cos(\theta-\alpha)$ : its coefficients are $R\cos\alpha$ on cosine and $R\sin\alpha$ on sine. Compare coefficients for the form requested. If both original coefficients are zero, the expression is identically zero and there is no uniquely determined phase angle.

Example 7: maximum on a full cycle and on a restricted interval

Write $3\sin\theta+4\cos\theta$ as $R\sin(\theta+\alpha)$ , with $R>0$ , $0^\circ<\alpha<90^\circ$ . Find its maximum for $0^\circ\le\theta\le360^\circ$ , then for $0^\circ\le\theta\le30^\circ$ .

Matching coefficients gives $R=5$ , $\cos\alpha=3/5$ and $\sin\alpha=4/5$ , so $\alpha=\tan^{-1}(4/3)\approx53.1301^\circ$ . The expression is $5\sin(\theta+\alpha)$ .

On the full cycle its maximum is $5$ , attained when $\theta+\alpha=90^\circ$ , namely $\theta=90^\circ-\alpha\approx36.9^\circ$ . On $[0^\circ,30^\circ]$ , the shifted angle stays in $[\alpha,\alpha+30^\circ]$ , entirely below 90°, where sine increases. Thus the restricted maximum occurs at $\theta=30^\circ$ :

3\sin30^\circ+4\cos30^\circ=\frac32+2\sqrt3.

The global bound 5 is not reached on that smaller interval. A maximum value and the angle at which it occurs are different parts of the answer.

Example 8: a reciprocal reverses the order

Find the range of $F(\theta)=\frac2{7+3\sin\theta+4\cos\theta}$ for $0^\circ\le\theta\le360^\circ$ .

The R-formula gives $-5\le3\sin\theta+4\cos\theta\le5$ , with both bounds attained. Hence the denominator lies in $[2,12]$ , so it is always positive. Taking reciprocals reverses their order:

\frac16\le F(\theta)\le1.

The minimum $1/6$ occurs when the sine combination is 5; the maximum 1 occurs when it is −5. If a denominator can cross zero, this positive-denominator argument cannot be used across the whole interval.

Explore: coefficients, amplitude and phase

Change a and b in $a\sin\theta+b\cos\theta$ . Compare the height of the resulting wave with $\sqrt{a^2+b^2}$ , and test what happens when a coefficient changes sign or both are zero.

Sine coefficient a:3Cosine coefficient b:4

Angles on this graph are in degrees, from 0° to 360°. Use the sliders or arrow keys.

The period is a full turn whenever at least one coefficient is nonzero. Changing the coefficients changes the amplitude and phase.

R = 5. Choose α ≈ 53.130°. The range is −5 to 5. At θ = 0°, the value is b = 4.

Common mistakes and useful checks

Assuming the identity to be proved. Transform one side into the other; do not begin with the desired equality and treat it as established.
Cancelling terms across addition. Factor first; only common nonzero factors can be cancelled.
Writing $\sin(A+B)=\sin A+\sin B$ . Use the products in the addition formula.
Mixing $\sin^2A$ with $\sin2A$ . The first squares a ratio; the second doubles an angle.
Forgetting a quadrant or a denominator restriction. List these before taking square roots or dividing.
Giving R as a maximum without checking the angle interval. Show that the required shifted angle can actually be reached.

Practise identities at the right level

For the first section, use the Secondary 3 basic trigonometric identities questions. For compound angles, double angles and R-formula, use the Secondary 4 trigonometric identities and R-formula questions. Apply these tools to trigonometric equations and graphs.