3.6 Trigonometric identities

The ratios $\sin\theta$, $\cos\theta$ and $\tan\theta$ are not independent of each other. They are coordinates and a gradient taken from the same point on the unit circle, so they are linked by equations that hold for every angle. These equations, called trigonometric identities, allow one ratio to be found from another, turn complicated expressions into simple ones and change an equation into a form that can be solved.

Identities are a clear example of equivalence: two expressions that look different but take the same value for every input. The same identities are used again to solve trigonometric equations, to differentiate and integrate the circular functions, and to model waves.

Identities and equations

An equation such as $\sin\theta = \frac{1}{2}$ is true only for particular values of $\theta$, which are found by solving it. An identity is true for every value of the variable for which both sides are defined. The symbol $\equiv$ ("is identically equal to") is sometimes used to stress this, as in $\tan\theta \equiv \frac{\sin\theta}{\cos\theta}$.

The difference matters for how each is handled. An equation is solved. An identity is proved, by showing that one side can be turned into the other using results already known (see 1.6). A single value of $\theta$ for which the two sides differ, a counterexample, is enough to show that a statement is not an identity.

The graphs of $y = \sin 2x$ and $y = 2\sin x$ make the same point: the curves are different, and they meet only at isolated points.

The tangent identity

The definition of tangent from 3.5 is the first identity:

It is in the formula booklet. It is used to replace $\tan\theta$ whenever an expression mixes tangent with sine and cosine, since a mixture of different ratios is usually easier to handle once everything is written in terms of $\sin\theta$ and $\cos\theta$.

The Pythagorean identity

Derivation from the unit circle

A point $P$ on the unit circle at angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$. Every point on the unit circle satisfies $x^2 + y^2 = 1$, because its distance from the origin is $1$. Substituting the coordinates of $P$ gives the Pythagorean identity.

For an acute angle this is Pythagoras's theorem in the right-angled triangle with hypotenuse $1$. The unit circle argument shows more: it holds in every quadrant, because the signs of $\cos\theta$ and $\sin\theta$ disappear when they are squared. The identity is in the formula booklet.

Rearranged forms

The identity is often used in one of its rearranged forms:

These replace one squared ratio by the other, which is how an expression or an equation containing both $\sin\theta$ and $\cos^2\theta$ is turned into one containing only $\sin\theta$.

Finding one ratio from another

When one ratio of an angle is known, the Pythagorean identity gives the square of another. Taking the square root gives two possible values, positive and negative, and the quadrant of the angle decides which sign is correct.

The method:

1. Use $\cos^2\theta + \sin^2\theta = 1$ to find the square of the unknown ratio.

2. Take the square root, keeping both signs.

3. Use the given quadrant (and CAST from 3.5) to choose the sign.

4. If $\tan\theta$ is needed, use $\tan\theta = \frac{\sin\theta}{\cos\theta}$.

The same result can be seen geometrically. The point at angle $\theta$ on a circle of radius $5$ has a $y$-coordinate of $3$, so it is $P(-4, 3)$, in the second quadrant.

When the given ratio is the tangent, it is often quicker to draw a right-angled triangle for the related angle than to use the identity, then attach the signs from CAST.

Questions can also give the ratio as a letter. The method is the same; the answer is an expression in that letter.

Double angle identities

The identities

The double angle identities express the ratios of $2\theta$ in terms of the ratios of $\theta$. Both are in the formula booklet.

Where the identities come from

For $0 < \theta < \frac{\pi}{2}$, both identities can be seen in one isosceles triangle. Take triangle $OAB$ with $OA = OB = 1$ and angle $2\theta$ at $O$. The altitude $OM$ bisects the angle at $O$ and the base $AB$, splitting the triangle into two right-angled triangles, each with hypotenuse $1$ and angle $\theta$ at $O$. In each, $AM = MB = \sin\theta$ and $OM = \cos\theta$.

Sine. The area of triangle $OAB$ can be found in two ways. Using $\frac{1}{2}ab\sin C$ with the two sides of length $1$ and the angle $2\theta$ between them gives $\frac{1}{2}\sin 2\theta$. Using $\frac{1}{2} \times \text{base} \times \text{height}$ gives $\frac{1}{2} \times 2\sin\theta \times \cos\theta = \sin\theta\cos\theta$. Setting these equal:

Cosine. The base $AB$ can also be found in two ways. The cosine rule in triangle $OAB$ gives $AB^2 = 1^2 + 1^2 - 2\cos 2\theta = 2 - 2\cos 2\theta$. From the diagram, $AB = 2\sin\theta$, so $AB^2 = 4\sin^2\theta$. Setting these equal:

This argument only covers acute $\theta$, but the identities hold for every angle. A general proof uses the compound angle identities for $\sin(A + B)$ and $\cos(A + B)$ with $A = B = \theta$, which HL students meet in 3.10.

The three forms of cos 2θ

The other two forms of $\cos 2\theta$ follow from $1 - 2\sin^2\theta$ by the Pythagorean identity. Replacing $1$ by $\cos^2\theta + \sin^2\theta$:

Replacing $\sin^2\theta$ by $1 - \cos^2\theta$:

Finding double angle ratios

Other multiples of the angle

"Double" is relative: the identities link any angle with its half. Replacing $\theta$ by $2x$, $3x$ or $\frac{x}{2}$ gives, for example,

Writing squares in terms of cos 2θ

Rearranging the two single-ratio forms of $\cos 2\theta$ gives expressions for $\cos^2\theta$ and $\sin^2\theta$ with no squares in them:

These are used to integrate $\sin^2 x$ and $\cos^2 x$ in calculus, and to find exact values for half of a standard angle. The graph shows that $y = \cos^2 x$ and $y = \frac{1}{2} + \frac{1}{2}\cos 2x$ are the same curve.

A double angle in a model

Double angles appear naturally in models. A ball launched with speed $v$ at an angle $\theta$ above horizontal ground, with air resistance ignored, lands at a horizontal distance

where $g \approx 9.8$ m s⁻² is the acceleration due to gravity. The double angle identity turns this into $R = \frac{v^2\sin 2\theta}{g}$, which shows at once how the range depends on the angle.

Proving identities

The method

To prove an identity, start from one side and transform it, step by step, into the other side. Each step uses algebra or an identity that is already known. The usual approach:

• Start with the more complicated side; there is more to simplify.

• Write $\tan\theta$ as $\frac{\sin\theta}{\cos\theta}$, so that only sines and cosines remain.

• Replace double angles by single angles (or the other way round) so that all the angles match.

• Look for $\cos^2\theta + \sin^2\theta$, which can be replaced by $1$, or for $1$, which can be replaced by $\cos^2\theta + \sin^2\theta$.

• Use algebra: expand brackets, factorise, add fractions over a common denominator, cancel common factors.

• End with "= RHS" (or "= LHS") so it is clear that the proof is complete.

Identities with fractions

When an identity contains a fraction with a double angle, choose the form of $\cos 2\theta$ that simplifies the numerator or the denominator. In $1 - \cos 2\theta$, the form $1 - 2\sin^2\theta$ makes the $1$s cancel; in $1 + \cos 2\theta$, the form $2\cos^2\theta - 1$ does the same.

When one side contains a sum of fractions, adding them over a common denominator usually produces $\sin^2\theta + \cos^2\theta$ or $1 - \sin^2\theta$ in the numerator or denominator.

Denominators of the form $1 - \sin\theta$ and $1 + \sin\theta$ multiply to a difference of two squares, $1 - \sin^2\theta$, which the Pythagorean identity turns into $\cos^2\theta$. The same happens with $1 \pm \cos\theta$.

Rewriting equations with identities

Identities are also used to rewrite an equation so that it contains only one trigonometric ratio of one angle, ready to be solved as in 3.8. Typical steps are replacing $\cos^2 x$ by $1 - \sin^2 x$ and replacing $\cos 2x$ by the form that matches the rest of the equation.

Exam-style questions on identities

Identity questions are usually set without a calculator and in several linked parts: find one ratio from another, use it in a double angle identity, then prove or apply a related result.

Self-check

• To do:

  I can explain why $\cos^2\theta + \sin^2\theta = 1$ for every angle, using the unit circle.

• To do:

  I can find $\sin\theta$, $\cos\theta$ or $\tan\theta$ from one other ratio, choosing the sign from the quadrant.

• To do:

  I can state $\sin 2\theta = 2\sin\theta\cos\theta$ and the three forms of $\cos 2\theta$, and choose the most useful form.

• To do:

  I can find the exact values of $\sin 2\theta$, $\cos 2\theta$ and $\tan 2\theta$ from one ratio of $\theta$.

• To do:

  I can recognise expressions such as $2\sin A\cos A$, $\cos^2 A - \sin^2 A$ and $1 - 2\sin^2 A$

• To do:

  I can prove an identity by working from one side to the other in a single chain of equalities.

• To do:

  I can use identities to rewrite an equation in terms of a single trigonometric ratio.

[6]6 marks

(c) Expand as a difference of two squares, then use the identity.

An obtuse angle lies in the second quadrant, where cosine is negative, so

Then

The angle is in the third quadrant, where sine is negative, so $\sin\theta = -\frac{12}{13}$.

This is positive, as expected: tangent is positive in the third quadrant.

The angle is in the fourth quadrant, where cosine is positive and sine is negative.

Check: $\frac{\sin\theta}{\cos\theta} = \frac{-2/\sqrt{5}}{1/\sqrt{5}} = -2$, and $\sin^2\theta + \cos^2\theta = \frac{4}{5} + \frac{1}{5} = 1$.

In the second quadrant cosine is negative, so

A check on the signs: $\theta \approx 143.1°$, so $2\theta \approx 286.3°$, in the fourth quadrant, where sine is negative and cosine is positive.

For $\sin 2\theta$, first find $\sin\theta$. In the first quadrant sine is positive:

(b) $\cos^2\theta - \sin^2\theta = \cos 2\theta$, so

(c) $1 - 2\sin^2\theta = \cos 2\theta$, so

The angle $\frac{\pi}{8}$ is in the first quadrant, so the cosine is positive.

A calculator check: both $\cos\frac{\pi}{8}$ and $\frac{\sqrt{2 + \sqrt{2}}}{2}$ equal $0.92388\ldots$.

(b) The largest value of $\sin 2\theta$ is $1$, when $2\theta = 90°$, so $\theta = 45°$.

(c) For $\theta = 30°$, $\sin 2\theta = \sin 60°$; for $\theta = 60°$, $\sin 2\theta = \sin 120° = \sin(180° - 60°) = \sin 60°$. The ranges are equal:

In general, complementary angles $\theta$ and $90° - \theta$ give the same range, since $\sin(180° - 2\theta) = \sin 2\theta$. The model is a simplification: with air resistance, the best angle is lower than $45°$.

Multiplying the numerator and denominator by $2$ in the last line creates the $2\sin\theta\cos\theta$ of the double angle identity.

Multiply by $-1$:

This is a quadratic equation in $\sin x$, which factorises as $(2\sin x + 1)(\sin x - 1) = 0$.

$x$ is in the second quadrant, where sine is positive, so $\sin x = \frac{\sqrt{5}}{3}$.

(b) $\sin 2x = 2\sin x\cos x = 2 \times \frac{\sqrt{5}}{3} \times \left(-\frac{2}{3}\right) = -\frac{4\sqrt{5}}{9}$

(c) Only $\cos x$ was given, so use the form in $\cos x$:

(d) $\tan 2x = \frac{\sin 2x}{\cos 2x} = \frac{-\frac{4\sqrt{5}}{9}}{-\frac{1}{9}} = 4\sqrt{5}$

(b) "Hence" means use part (a) with $\theta = \frac{\pi}{8}$, so that $2\theta = \frac{\pi}{4}$.

Rationalise the denominator by multiplying by $\frac{2 - \sqrt{2}}{2 - \sqrt{2}}$:

So $\tan\frac{\pi}{8} = -1 + \sqrt{2}$, with $a = -1$ and $b = 1$.