IB Analysis and Approaches HL · Trigonometry · Section 3 of 5

Sum, double-angle and half-angle identities

3 worked examples · By Álvaro López

Key idea

With the sum formulas for two angles you can obtain all the others: the double angle, the half angle, and exact values such as $sin 7 5^{\circ}$ , which are not in the table of $3 0^{\circ}$ , $4 5^{\circ}$ and $6 0^{\circ}$ . They also serve to simplify expressions and to prove identities, a very typical type of question in Paper 1.

Sine, cosine and tangent of a sum and of a difference

It is tempting to think that $sin (A + B)$ is $sin A + sin B$ , but it is not: with $A = B = 9 0^{\circ}$ one side would give $sin 18 0^{\circ} = 0$ and the other $1 + 1 = 2$ . The sine of a sum mixes the sines and cosines of the two angles.

Note the sign of the cosine: it is the opposite of the one for the sum. It can be understood with $cos (A + B)$ when $A$ and $B$ are small: the cosine of a larger angle must be smaller, and that is why the product $sin A sin B$ is subtracted.

Where it all comes from: the formula for the sine of a sum can be proved with a diagram on the unit circle, and the others are deduced from it. For the difference, just replace $B$ by $- B$ and recall that $sin (- B) = - sin B$ and $cos (- B) = cos B$ . The cosine one comes from $cos x = sin (9 0^{\circ} - x)$ , and the tangent one from dividing sine by cosine:

$tan (A + B) = \frac{sin A cos B + cos A sin B}{cos A cos B - sin A sin B} = \frac{tan A + tan B}{1 - tan A tan B}$

(divide numerator and denominator by $cos A cos B$ ).

Exact values of new angles

Using the sum formulas you can obtain the exact ratios of any angle that can be written as a sum or difference of angles with known ratios (those in the table of exact values and their equivalents in other quadrants, such as $12 0^{\circ}$ or $21 0^{\circ}$ ). For example, $7 5^{\circ} = 4 5^{\circ} + 3 0^{\circ}$ and $1 5^{\circ} = 6 0^{\circ} - 4 5^{\circ}$ .

Double angle

If in the sum formulas you take $B = A$ , you obtain the double-angle formulas. Use the letter $θ$ for the angle:

There are three forms of the cosine. The second and third come from the first by substituting $sin^{2} θ = 1 - cos^{2} θ$ or $cos^{2} θ = 1 - sin^{2} θ$ (from $sin^{2} θ + cos^{2} θ = 1$ ). Which one to use depends on what you know: if you only know $cos θ$ , use $2 cos^{2} θ - 1$ ; if you only know $sin θ$ , use $1 - 2 sin^{2} θ$ .

A very frequent use is solving for the squares. From the last two forms one obtains

$cos^{2} θ = \frac{1 + cos 2 θ}{2} sin^{2} θ = \frac{1 - cos 2 θ}{2}$

which allow you to turn a square of sine or cosine into a first-degree expression (this is used a great deal in integrals and in equations).

Half angle

The half angle is the double-angle formula read backwards. If in $cos 2 α = 2 cos^{2} α - 1$ you call $θ = 2 α$ , that is $α = \frac{θ}{2}$ , you get

$cos θ = 2 cos^{2} \frac{θ}{2} - 1 and, with the other form, cos θ = 1 - 2 sin^{2} \frac{θ}{2}$

These formulas are not usually given in the IB formula booklet in this form: the usual thing is to derive them from the cosine of the double angle, as you have just seen. It is worth knowing how to do so.

Proving identities

An identity is an equality that holds for all values of the variable for which both sides exist, such as $sin^{2} x + cos^{2} x = 1$ . Proving an identity means showing that the left-hand side and the right-hand side are the same expression.

Simplifying and calculating with identities

Identities also serve to reduce an expression to a simpler one, or to calculate values. Two useful recognitions:

$sin A cos B + cos A sin B$ is $sin (A + B)$ , and $cos A cos B - sin A sin B$ is $cos (A + B)$ ; so $cos 4 7^{\circ} cos 1 3^{\circ} - sin 4 7^{\circ} sin 1 3^{\circ} = cos 6 0^{\circ} = \frac{1}{2}$ .
$2 sin θ cos θ$ is $sin 2 θ$ , and $cos^{2} θ - sin^{2} θ$ is $cos 2 θ$ ; so $2 sin 1 5^{\circ} cos 1 5^{\circ} = sin 3 0^{\circ} = \frac{1}{2}$ .

When you are given a ratio of $θ$ and the quadrant, first calculate the other ratios of $θ$ with $sin^{2} θ + cos^{2} θ = 1$ (choosing the sign by the quadrant), and then apply the formula that corresponds.

Worked examples

Example 1 Basic

Exact values with sum and difference

Calculate the exact value of:

a) $sin 7 5^{\circ}$

b) $cos 16 5^{\circ}$ (write $16 5^{\circ} = 12 0^{\circ} + 4 5^{\circ}$ )

c) $tan 7 5^{\circ}$

d) $sin 7 5^{\circ} cos 1 5^{\circ} + cos 7 5^{\circ} sin 1 5^{\circ}$

Solution

a) With $7 5^{\circ} = 4 5^{\circ} + 3 0^{\circ}$ :

$sin 7 5^{\circ} = sin 4 5^{\circ} cos 3 0^{\circ} + cos 4 5^{\circ} sin 3 0^{\circ} = \frac{2}{2} \cdot \frac{3}{2} + \frac{2}{2} \cdot \frac{1}{2} = \frac{6 + 2}{4}$

b) With $16 5^{\circ} = 12 0^{\circ} + 4 5^{\circ}$ , and noting that $cos 12 0^{\circ} = - \frac{1}{2}$ and $sin 12 0^{\circ} = \frac{3}{2}$ :

$cos 16 5^{\circ} = cos 12 0^{\circ} cos 4 5^{\circ} - sin 12 0^{\circ} sin 4 5^{\circ} = - \frac{1}{2} \cdot \frac{2}{2} - \frac{3}{2} \cdot \frac{2}{2} = - \frac{2 + 6}{4}$

The negative sign is consistent: $16 5^{\circ}$ lies in the second quadrant.

c) With $7 5^{\circ} = 4 5^{\circ} + 3 0^{\circ}$ , $tan 4 5^{\circ} = 1$ and $tan 3 0^{\circ} = \frac{1}{3} = \frac{3}{3}$ :

$tan 7 5^{\circ} = \frac{1 + \frac{3}{3}}{1 - \frac{3}{3}} = \frac{3 + 3}{3 - 3} = \frac{( 3 + 3 ) ^{2}}{9 - 3} = \frac{12 + 6 3}{6} = 2 + 3$

d) It is the expansion of $sin (7 5^{\circ} + 1 5^{\circ})$ , so it equals $sin 9 0^{\circ} = 1$ .

Example 2 Medium

Double and half angle from the sine and the quadrant

The angle $θ$ satisfies $sin θ = \frac{5}{13}$ and $9 0^{\circ} < θ < 18 0^{\circ}$ . The angle $φ$ satisfies $cos φ = \frac{4}{5}$ and $27 0^{\circ} < φ < 36 0^{\circ}$ .

a) Find $cos θ$ , $sin 2 θ$ and $cos 2 θ$ .

b) Find $cos \frac{θ}{2}$ and $tan \frac{θ}{2}$ .

c) Find $sin (θ + φ)$ .

Solution

a) From $sin^{2} θ + cos^{2} θ = 1$ we have $cos^{2} θ = 1 - \frac{25}{169} = \frac{144}{169}$ , that is, $cos θ = \pm \frac{12}{13}$ . As $θ$ lies in the second quadrant, the cosine is negative: $cos θ = - \frac{12}{13}$ .

Using the double-angle formulas:

$sin 2 θ = 2 sin θ cos θ = 2 \cdot \frac{5}{13} \cdot (- \frac{12}{13}) = - \frac{120}{169}$

$cos 2 θ = 1 - 2 sin^{2} θ = 1 - 2 \cdot \frac{25}{169} = \frac{119}{169}$

(Check with the other form: $cos^{2} θ - sin^{2} θ = \frac{144 - 25}{169} = \frac{119}{169}$ .)

b) As $9 0^{\circ} < θ < 18 0^{\circ}$ , the half angle satisfies $4 5^{\circ} < \frac{θ}{2} < 9 0^{\circ}$ : it lies in the first quadrant, so its cosine is positive.

$cos \frac{θ}{2} = \frac{1 + cos θ}{2} = \frac{1 - \frac{12}{13}}{2} = \frac{1}{26} = \frac{26}{26}$

For the tangent the sign is not needed:

$tan \frac{θ}{2} = \frac{sin θ}{1 + cos θ} = \frac{\frac{5}{13}}{1 - \frac{12}{13}} = \frac{5/13}{1/13} = 5$

c) For $φ$ in the fourth quadrant the sine is negative: $sin^{2} φ = 1 - \frac{16}{25} = \frac{9}{25}$ , so $sin φ = - \frac{3}{5}$ . Then

$sin (θ + φ) = sin θ cos φ + cos θ sin φ = \frac{5}{13} \cdot \frac{4}{5} + (- \frac{12}{13}) (- \frac{3}{5}) = \frac{20 + 36}{65} = \frac{56}{65}$

Example 3 HardProof

Prove three identities

Prove the following identities:

a) $sin 3 θ = 3 sin θ - 4 sin^{3} θ$

b) $tan x + \frac{1}{tan x} = \frac{2}{sin 2 x}$

c) $cos (A + B) cos (A - B) = cos^{2} A - sin^{2} B$

Solution

a) We work only with the left-hand side, writing $3 θ = 2 θ + θ$ and using the double angle:

$sin 3 θ = sin (2 θ + θ) = sin 2 θ cos θ + cos 2 θ sin θ = 2 sin θ cos^{2} θ + (1 - 2 sin^{2} θ) sin θ = 2 sin θ (1 - sin^{2} θ) + sin θ - 2 sin^{3} θ = 2 sin θ - 2 sin^{3} θ + sin θ - 2 sin^{3} θ = 3 sin θ - 4 sin^{3} θ$

We have chosen $cos 2 θ = 1 - 2 sin^{2} θ$ and $cos^{2} θ = 1 - sin^{2} θ$ so that everything is in terms of $sin θ$ , which is what appears on the right-hand side.

b) We start from the left-hand side and express everything in sines and cosines (for $x$ with $sin x \neq = 0$ and $cos x \neq = 0$ ):

$tan x + \frac{1}{tan x} = \frac{sin x}{cos x} + \frac{cos x}{sin x} = \frac{sin ^{2} x + cos ^{2} x}{sin x cos x} = \frac{1}{sin x cos x} = \frac{2}{2 sin x cos x} = \frac{2}{sin 2 x}$

c) We expand the left-hand side with the sum and difference formulas. It is a product of the form $(P - Q) (P + Q)$ with $P = cos A cos B$ and $Q = sin A sin B$ :

$cos (A + B) cos (A - B) = (cos A cos B - sin A sin B) (cos A cos B + sin A sin B) = cos^{2} A cos^{2} B - sin^{2} A sin^{2} B = cos^{2} A (1 - sin^{2} B) - (1 - cos^{2} A) sin^{2} B = cos^{2} A - cos^{2} A sin^{2} B - sin^{2} B + cos^{2} A sin^{2} B = cos^{2} A - sin^{2} B$

In the exam: a proof starts from one side and ends at the other, without writing "left-hand side $=$ right-hand side" at any step before it has been shown. Mark each transformation as well justified (for example, "by the double-angle formula") and give the final exact expression that the question asks for.

Common mistakes

Summary

Sum and difference: $sin (A \pm B) = sin A cos B \pm cos A sin B$ ; $cos (A \pm B) = cos A cos B \mp sin A sin B$ ; $tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B}$ .
They give exact values such as $sin 7 5^{\circ} = \frac{6 + 2}{4}$ and $tan 7 5^{\circ} = 2 + 3$ , by writing the angle as a sum or difference of known angles.
Double angle: $sin 2 θ = 2 sin θ cos θ$ ; $cos 2 θ = cos^{2} θ - sin^{2} θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ$ ; $tan 2 θ = \frac{2 tan θ}{1 - tan ^{2} θ}$ .
Half angle (derived from the double angle): $cos \frac{θ}{2} = \pm \frac{1 + c o s θ}{2}$ , $sin \frac{θ}{2} = \pm \frac{1 - c o s θ}{2}$ , $tan \frac{θ}{2} = \frac{s i n θ}{1 + c o s θ}$ ; the sign depends on the quadrant of $\frac{θ}{2}$ .
Given a ratio and the quadrant, first calculate the other ratios of $θ$ with $sin^{2} θ + cos^{2} θ = 1$ and choose the sign.
To prove an identity: start from one side only, transform with justified steps, and arrive exactly at the other; do not operate on both sides at once.

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