Trigonometric Formulae, Equations and Inequalities

Lessons

Senior 5 · Mathematics

Trigonometric Formulae, Equations and Inequalities

5 lessons

Lesson 2 of 5

Double-angle, half-angle and product–sum formulae

25 min

Put $B = A$

The addition formulae of Lesson 1 hold for every pair of angles, so they hold when the two angles are the same. Setting $B = A$ in $\sin(A + B)$ , $\cos(A + B)$ and $\tan(A + B)$ gives the double-angle formulae:

\sin 2A = \sin A\cos A + \cos A\sin A = 2\sin A\cos A

\cos 2A = \cos^2 A - \sin^2 A

\tan 2A = \frac{2\tan A}{1 - \tan^2 A} \quad \left(\tan A \ne \pm 1,\ \cos A \ne 0\right)

Replacing $\sin^2 A$ by $1 - \cos^2 A$ , or $\cos^2 A$ by $1 - \sin^2 A$ , gives two more forms of $\cos 2A$ . Choose the form that suits the question:

\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A.

Definition

Double-angle formulae: identities that give $\sin 2A$ , $\cos 2A$ and $\tan 2A$ in terms of the ratios of $A$ .

Half-angle (power-reducing) formulae: identities that give the ratios of $\frac{\theta}{2}$ in terms of $\cos\theta$ , or that replace a square such as $\cos^2 A$ by a first power of $\cos 2A$ .

Example: $\sin 60^\circ = 2\sin 30^\circ\cos 30^\circ = 2\cdot\frac{1}{2}\cdot\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$ .

Worked example 1: a third-quadrant angle

$\cos A = -\frac{7}{25}$ and $\pi \lt A \lt \frac{3\pi}{2}$ . Find $\sin 2A$ , $\cos 2A$ and $\tan 2A$ , and say in which quadrant $2 A$ lies.

Solution In the third quadrant the sine is negative: $\sin A = -\sqrt{1 - \frac{49}{625}} = -\frac{24}{25}$ .

\sin 2A = 2\left(-\frac{24}{25}\right)\left(-\frac{7}{25}\right) = \frac{336}{625}, \qquad \cos 2A = \frac{49}{625} - \frac{576}{625} = -\frac{527}{625}, \qquad \tan 2A = -\frac{336}{527}.

$\sin 2A \gt 0$ and $\cos 2A \lt 0$ , so $2 A$ is in the second quadrant (it lies between $2\pi$ and $3\pi$ , one full turn plus a second-quadrant angle). Check: $336^2 + 527^2 = 390\,625 = 625^2$ . ✓

Half-angle formulae

Solve the last two forms of $\cos 2A$ for the squares:

\cos^2 A = \frac{1 + \cos 2A}{2}, \qquad \sin^2 A = \frac{1 - \cos 2A}{2}.

These power-reducing forms turn a square into a first power of a double angle; you will use them again for integration in S6. Now write $A = \frac{\theta}{2}$ and take square roots:

\cos\frac{\theta}{2} = \pm\sqrt{\frac{1 + \cos\theta}{2}}, \qquad \sin\frac{\theta}{2} = \pm\sqrt{\frac{1 - \cos\theta}{2}}, \qquad \tan\frac{\theta}{2} = \frac{\sin\theta}{1 + \cos\theta} = \frac{1 - \cos\theta}{\sin\theta}.

The sign is the sign of the ratio in the quadrant where $\frac{\theta}{2}$ lies. The tangent forms need no sign choice: multiply $\tan\frac{\theta}{2} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}$ above and below by $2\cos\frac{\theta}{2}$ to get $\frac{\sin\theta}{1 + \cos\theta}$ .

Figure 1.

\sin^2 x = \frac{1}{2} - \frac{1}{2}\cos 2x

: a cosine wave of amplitude

\frac{1}{2}

and period

π \pi

, turned upside down and lifted to the line

y = \frac{1}{2}

The t-formulae. With $t = \tan\frac{x}{2}$ , divide $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ and $\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$ by $1 = \cos^2\frac{x}{2} + \sin^2\frac{x}{2}$ , then divide the top and the bottom by $\cos^2\frac{x}{2}$ :

\sin x = \frac{2t}{1 + t^2}, \qquad \cos x = \frac{1 - t^2}{1 + t^2}, \qquad \tan x = \frac{2t}{1 - t^2}.

Every ratio of $x$ becomes a fraction in one letter $t$ . Lesson 4 uses this to solve equations.

Worked example 2: exact values for $\frac{\pi}{8}$

Find $\cos\frac{\pi}{8}$ and $\tan\frac{\pi}{8}$ exactly.

Solution $\frac{\pi}{8}$ is half of $\frac{\pi}{4}$ , and it is in the first quadrant, so the cosine is positive:

\cos\frac{\pi}{8} = \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{2}}{4}} = \frac{\sqrt{2 + \sqrt{2}}}{2} \approx 0.9239,

\tan\frac{\pi}{8} = \frac{1 - \cos\frac{\pi}{4}}{\sin\frac{\pi}{4}} = \frac{1 - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2 - \sqrt{2}}{\sqrt{2}} = \sqrt{2} - 1 \approx 0.4142.

Worked example 3: a triple angle

Express $\cos 3x$ in terms of $\cos x$ only.

Solution Write $3 x = 2 x + x$ and use the addition formula, then the double-angle formulae:

\cos 3x = \cos 2x\cos x - \sin 2x\sin x = (2\cos^2 x - 1)\cos x - 2\sin^2 x\cos x

= 2\cos^3 x - \cos x - 2(1 - \cos^2 x)\cos x = 4\cos^3 x - 3\cos x.

Check with $x = \frac{\pi}{3}$ : $\cos\pi = -1$ and $4\left(\frac{1}{2}\right)^3 - 3\cdot\frac{1}{2} = \frac{1}{2} - \frac{3}{2} = -1$ . ✓

Warning

$\sin 2A$ is not $2\sin A$ . For $A = 60^\circ$ : $\sin 120^\circ = \frac{\sqrt{3}}{2} \approx 0.87$ , but $2\sin 60^\circ = \sqrt{3} \approx 1.73$ . In the half-angle formulae, do not drop the $\pm$ : decide the sign from the quadrant of $\frac{\theta}{2}$ , not of $\theta$ .

Products into sums, and sums into products

Add and subtract the expansions of $\sin(A + B)$ and $\sin(A - B)$ , and of $\cos(A + B)$ and $\cos(A - B)$ . The unwanted terms cancel:

Product into sum	Sum into product (with $p = A + B$ , $q = A - B$ )
$2\sin A\cos B = \sin(A + B) + \sin(A - B)$	$\sin p + \sin q = 2\sin\frac{p + q}{2}\cos\frac{p - q}{2}$
$2\cos A\sin B = \sin(A + B) - \sin(A - B)$	$\sin p - \sin q = 2\cos\frac{p + q}{2}\sin\frac{p - q}{2}$
$2\cos A\cos B = \cos(A + B) + \cos(A - B)$	$\cos p + \cos q = 2\cos\frac{p + q}{2}\cos\frac{p - q}{2}$
$2\sin A\sin B = \cos(A - B) - \cos(A + B)$	$\cos p - \cos q = -2\sin\frac{p + q}{2}\sin\frac{p - q}{2}$

The right-hand column is the left-hand column read backwards: if $p = A + B$ and $q = A - B$ , then $A = \frac{p + q}{2}$ and $B = \frac{p - q}{2}$ . Products are easier to integrate when written as sums; sums are easier to solve (set to zero) when written as products.

Figure 2.

\sin 9x + \sin 7x = 2\sin 8x\cos x

. The fast wave

\sin 8x

is kept inside the slow envelope

\pm 2\cos x

. Two musical notes of close frequencies combine in the same way, and you hear the loudness rise and fall: these are called beats.

Definition

Product-to-sum (and sum-to-product) formulae: identities that rewrite a product of two sines or cosines as a sum or difference of sines or cosines, and the other way round.

Example: $2\sin 5x\cos 3x = \sin 8x + \sin 2x$ , and $\sin 8x + \sin 2x = 2\sin 5x\cos 3x$ .

Worked example 4: using both directions

(a) Find $\cos 75^\circ\cos 15^\circ$ exactly. (b) Simplify $\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x}$ .

Solution

$\cos 75^\circ\cos 15^\circ = \frac{1}{2}\left[\cos 90^\circ + \cos 60^\circ\right] = \frac{1}{2}\left(0 + \frac{1}{2}\right) = \frac{1}{4}$ .
Top: $\sin 5x + \sin 3x = 2\sin 4x\cos x$ . Bottom: $\cos 5x + \cos 3x = 2\cos 4x\cos x$ . So, where $\cos x \ne 0$ and $\cos 4x \ne 0$ , the fraction is $\frac{2\sin 4x\cos x}{2\cos 4x\cos x} = \tan 4x$ .

Tip

Remember the pattern, not twelve lines. In "sum into product" the angles are always the half-sum $\frac{p + q}{2}$ and the half-difference $\frac{p - q}{2}$ . Check a formula you are unsure of with one quick number, such as $p = q$ .

Mathematical Visual Proofs: A proof to remember: Double Angle Formulas I (about 3 min, no narration). Watch how one isosceles triangle with apex angle 2θ gives sin 2θ and cos 2θ from the sine and cosine rules you learned in S4.

✦

Practice

Exact answers, no calculator. The last two are harder.

$\sin A = \frac{3}{5}$ and $A$ is acute. Find $\sin 2A$ , $\cos 2A$ and $\tan 2A$ .
Simplify (a) $2\sin 3x\cos 3x$ (b) $\cos^2 4x - \sin^2 4x$ (c) $1 - 2\sin^2\frac{\pi}{12}$ .
Find $\sin 22.5^\circ$ exactly.
Write $\sin^4 x$ in terms of $\cos 2x$ and $\cos 4x$ .
Write as a sum or difference: (a) $2\cos 4x\cos x$ (b) $2\sin 3x\sin x$ .
Write as a product: (a) $\sin 7x + \sin 3x$ (b) $\cos x - \cos 5x$ .
Prove that $\frac{\sin 2x}{1 + \cos 2x} = \tan x$ , and state when it holds.
Prove that $\cos 20^\circ\cos 40^\circ\cos 80^\circ = \frac{1}{8}$ . (Hint: multiply by $\sin 20^\circ$ and use $\sin 2A$ three times.)

Show answersHide answers

$\cos A = \frac{4}{5}$ . $\sin 2A = \frac{24}{25}$ , $\cos 2A = \frac{16 - 9}{25} = \frac{7}{25}$ , $\tan 2A = \frac{24}{7}$ .
(a) $\sin 6x$ (b) $\cos 8x$ (c) $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ .
$\sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{2}}}{2} \approx 0.3827$ .
$\sin^4 x = \left(\frac{1 - \cos 2x}{2}\right)^2 = \frac{1 - 2\cos 2x + \cos^2 2x}{4} = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$ .
(a) $\cos 5x + \cos 3x$ (b) $\cos 2x - \cos 4x$ .
(a) $2\sin 5x\cos 2x$ (b) $-2\sin 3x\sin(-2x) = 2\sin 3x\sin 2x$ .
$\frac{2\sin x\cos x}{2\cos^2 x} = \frac{\sin x}{\cos x} = \tan x$ , valid where $\cos x \ne 0$ (then $1 + \cos 2x = 2\cos^2 x \ne 0$ ).
$\sin 20^\circ\cos 20^\circ\cos 40^\circ\cos 80^\circ = \frac{1}{2}\sin 40^\circ\cos 40^\circ\cos 80^\circ = \frac{1}{4}\sin 80^\circ\cos 80^\circ = \frac{1}{8}\sin 160^\circ = \frac{1}{8}\sin 20^\circ$ . Divide by $\sin 20^\circ \ne 0$ .

Sources and credits

Written by Ganzaa: the explanations, the derivations, the graphs, the worked examples, the practice items and the quiz questions.
Video: A proof to remember: Double Angle Formulas I (visual proof), Mathematical Visual Proofs, embedded from YouTube.

1Put B=A B = A B=A

2Half-angle formulae

3Products into sums, and sums into products

Put $B = A$

Half-angle formulae

Products into sums, and sums into products