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Trigonometric identities and equations

Algebra · level 8, lesson 7 of 8

A question to start

You know \(\sin 30° = \tfrac12\) and \(\sin 45° = \tfrac{\sqrt2}{2}\) exactly. What is \(\sin 75°\) exactly?

It is tempting to add: \(75 = 45 + 30\), so maybe \(\sin 75° = \sin 45° + \sin 30° \approx 0.707 + 0.5 = 1.207\). But no sine is ever bigger than 1, so that cannot be right. The true value is

\[\sin 75° = \frac{\sqrt6 + \sqrt2}{4} \approx 0.9659.\]

Where does that come from? From a formula for \(\sin(a + b)\), which this lesson proves with nothing more than the unit circle and the distance formula.

The idea

What you need from geometry

On the unit circle (radius 1, centre the origin), the point at angle \(\theta\) is \((\cos\theta, \sin\theta)\). Angles are measured anticlockwise from the positive \(x\)-axis, in degrees or in radians (\(\pi\) radians \(= 180°\)). This gives three facts you will use:

  • \(\sin^2\theta + \cos^2\theta = 1\) (the point is at distance 1 from the origin);
  • \(\cos(-\theta) = \cos\theta\) and \(\sin(-\theta) = -\sin\theta\) (turning the other way reflects the point in the \(x\)-axis);
  • \(\sin\theta = \cos\left(\tfrac{\pi}{2} - \theta\right)\) and \(\cos\theta = \sin\left(\tfrac{\pi}{2} - \theta\right)\) (the cofunction identities).

All of this is in Radians and the unit circle.

An identity is an equation that is true for every value of the variable (where both sides are defined), like \(\sin^2\theta + \cos^2\theta = 1\). An ordinary equation like \(2\sin x = 1\) is true only for some values, which we have to find.

The compound angle formula for \(\cos(a - b)\), proved

a − b P (cos a, sin a) Q (cos b, sin b) PQ 1

Put two points on the unit circle: \(P = (\cos a, \sin a)\) at angle \(a\) and \(Q = (\cos b, \sin b)\) at angle \(b\). The angle between them, at the centre, is \(a - b\).

Measure \(PQ\) directly. By the distance formula,

\[PQ^2 = (\cos a - \cos b)^2 + (\sin a - \sin b)^2.\]

Expand both squares and group \(\cos^2 a + \sin^2 a = 1\) and \(\cos^2 b + \sin^2 b = 1\):

\[PQ^2 = 2 - 2(\cos a\cos b + \sin a\sin b).\]

Measure it after a rotation. Turn the whole picture clockwise by \(b\). Turning does not change lengths. Now \(Q\) sits at \((1, 0)\) and \(P\) sits at angle \(a - b\), that is at \((\cos(a - b), \sin(a - b))\). So

\[PQ^2 = (\cos(a - b) - 1)^2 + \sin^2(a - b) = \cos^2(a - b) - 2\cos(a - b) + 1 + \sin^2(a - b) = 2 - 2\cos(a - b).\]

Compare. The two expressions for \(PQ^2\) must be equal, so

\[\cos(a - b) = \cos a\cos b + \sin a\sin b. \qquad \blacksquare\]

The other compound angle formulas

Everything else follows from this one with a little algebra.

\(\cos(a + b)\). Replace \(b\) by \(-b\), and use \(\cos(-b) = \cos b\) and \(\sin(-b) = -\sin b\):

\[\cos(a + b) = \cos a\cos b - \sin a\sin b.\]

\(\sin(a + b)\). Use the cofunction identity, then the formula for \(\cos\) of a difference:

\[\sin(a + b) = \cos\left(\left(\tfrac{\pi}{2} - a\right) - b\right) = \cos\left(\tfrac{\pi}{2} - a\right)\cos b + \sin\left(\tfrac{\pi}{2} - a\right)\sin b,\]

and since \(\cos\left(\tfrac{\pi}{2} - a\right) = \sin a\) and \(\sin\left(\tfrac{\pi}{2} - a\right) = \cos a\):

\[\sin(a + b) = \sin a\cos b + \cos a\sin b.\]

\(\sin(a - b)\). Replace \(b\) by \(-b\): \(\sin(a - b) = \sin a\cos b - \cos a\sin b\).

Double angle formulas

Put \(b = a\) in the sum formulas:

\[\sin 2a = 2\sin a\cos a,\]
\[\cos 2a = \cos^2 a - \sin^2 a = 2\cos^2 a - 1 = 1 - 2\sin^2 a.\]

The last two forms come from replacing \(\sin^2 a\) by \(1 - \cos^2 a\), or \(\cos^2 a\) by \(1 - \sin^2 a\). Having three forms of \(\cos 2a\) is useful: choose the one that matches the rest of your equation.

How to prove an identity

Start with one side (usually the more complicated one) and change it, step by step, into the other side, using known identities. Do not start by writing the identity and doing the same thing to both sides: that assumes what you are trying to prove. Useful moves: write everything in \(\sin\) and \(\cos\); use the Pythagorean identity; use common denominators; factor.

How to solve a trigonometric equation

  1. Use identities so that only one trig function of one angle appears.
  2. Solve for that function, like an ordinary equation (it is often a quadratic in disguise: put \(s = \sin x\)).
  3. Find every angle in the interval with that value, using the unit circle: most values come up twice per turn.
  4. Never divide by \(\sin x\) or \(\cos x\) (it might be 0); factor instead.

Worked examples

1. The opening question. Find \(\sin 75°\) exactly.

  • \(\sin(45° + 30°) = \sin 45°\cos 30° + \cos 45°\sin 30° = \dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2} + \dfrac{\sqrt2}{2}\cdot\dfrac12 = \dfrac{\sqrt6 + \sqrt2}{4} \approx 0.9659\). ✓

2. Proving an identity. Prove \((\sin x + \cos x)^2 = 1 + \sin 2x\).

  • Left side \(= \sin^2 x + 2\sin x\cos x + \cos^2 x = (\sin^2 x + \cos^2 x) + 2\sin x\cos x = 1 + \sin 2x\), the right side. ∎

3. Another identity. Prove \(\dfrac{1 - \cos 2x}{\sin 2x} = \tan x\).

  • Use \(\cos 2x = 1 - 2\sin^2 x\) on top and \(\sin 2x = 2\sin x\cos x\) below.
  • Left side \(= \dfrac{1 - (1 - 2\sin^2 x)}{2\sin x\cos x} = \dfrac{2\sin^2 x}{2\sin x\cos x} = \dfrac{\sin x}{\cos x} = \tan x\). ∎ (Valid where the left side is defined.)

4. A quadratic in \(\sin x\). Solve \(2\sin^2 x - \sin x - 1 = 0\) for \(0 \le x \le 2\pi\).

  • Put \(s = \sin x\): \(2s^2 - s - 1 = 0\), so \((2s + 1)(s - 1) = 0\), so \(s = 1\) or \(s = -\tfrac12\).
  • \(\sin x = 1\): \(x = \dfrac{\pi}{2}\).
  • \(\sin x = -\tfrac12\): sine is negative in the third and fourth quarters, with reference angle \(\tfrac{\pi}{6}\): \(x = \pi + \tfrac{\pi}{6} = \dfrac{7\pi}{6}\) and \(x = 2\pi - \tfrac{\pi}{6} = \dfrac{11\pi}{6}\).
  • Three solutions: \(\dfrac{\pi}{2}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}\).
x y π/2 π 3π/2 2π −1 1 2 7π/6 11π/6

5. Using a double angle. Solve \(\cos 2x = \sin x\) for \(0 \le x \le 2\pi\).

  • Choose the form of \(\cos 2x\) with only sines: \(1 - 2\sin^2 x = \sin x\), so \(2\sin^2 x + \sin x - 1 = 0\).
  • \((2\sin x - 1)(\sin x + 1) = 0\), so \(\sin x = \tfrac12\) or \(\sin x = -1\).
  • \(x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}\).

6. Do not divide. Solve \(\sin 2x = \sin x\) for \(0 \le x \le 2\pi\).

  • \(2\sin x\cos x - \sin x = 0\), so \(\sin x\,(2\cos x - 1) = 0\).
  • \(\sin x = 0\): \(x = 0, \pi, 2\pi\). \(\cos x = \tfrac12\): \(x = \dfrac{\pi}{3}, \dfrac{5\pi}{3}\).
  • Five solutions. Dividing by \(\sin x\) at the start would have lost \(0, \pi, 2\pi\).

Try it

  1. Find \(\cos 75°\) exactly, using \(75° = 45° + 30°\).
  2. Find \(\sin\dfrac{\pi}{12}\) exactly, using \(\dfrac{\pi}{12} = \dfrac{\pi}{4} - \dfrac{\pi}{6}\).
  3. \(\sin a = \tfrac35\) and \(a\) is between 0 and \(\tfrac{\pi}{2}\). Find \(\sin 2a\) and \(\cos 2a\).
  4. Prove \(\tan x + \dfrac{1}{\tan x} = \dfrac{2}{\sin 2x}\).
  5. Solve \(2\cos x + \sqrt3 = 0\) for \(0 \le x \le 2\pi\).
  6. Solve \(2\cos^2 x - \cos x = 0\) for \(0 \le x \le 2\pi\).
  7. Solve \(\sin^2 x = 3\cos^2 x\) for \(0 \le x \le 2\pi\).
  8. Prove \(\sin 3x = 3\sin x - 4\sin^3 x\).
Answers
  1. \(\cos 45°\cos 30° - \sin 45°\sin 30° = \dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2} - \dfrac{\sqrt2}{2}\cdot\dfrac12 = \dfrac{\sqrt6 - \sqrt2}{4} \approx 0.2588\).
  2. \(\sin\dfrac{\pi}{4}\cos\dfrac{\pi}{6} - \cos\dfrac{\pi}{4}\sin\dfrac{\pi}{6} = \dfrac{\sqrt6}{4} - \dfrac{\sqrt2}{4} = \dfrac{\sqrt6 - \sqrt2}{4}\). (The same as \(\cos 75°\), because \(15° + 75° = 90°\): a cofunction pair.)
  3. \(\cos a = \sqrt{1 - \tfrac{9}{25}} = \tfrac45\) (positive, since \(a\) is in the first quarter). \(\sin 2a = 2 \cdot \tfrac35 \cdot \tfrac45 = \tfrac{24}{25}\). \(\cos 2a = 1 - 2 \cdot \tfrac{9}{25} = \tfrac{7}{25}\). Check: \(\left(\tfrac{24}{25}\right)^2 + \left(\tfrac{7}{25}\right)^2 = \tfrac{576 + 49}{625} = 1\). ✓
  4. Left side \(= \dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} = \dfrac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \dfrac{1}{\sin x\cos x} = \dfrac{2}{2\sin x\cos x} = \dfrac{2}{\sin 2x}\). ∎
  5. \(\cos x = -\tfrac{\sqrt3}{2}\): cosine is negative in the second and third quarters, reference angle \(\tfrac{\pi}{6}\): \(x = \dfrac{5\pi}{6}\) and \(x = \dfrac{7\pi}{6}\).
  6. \(\cos x\,(2\cos x - 1) = 0\). \(\cos x = 0\): \(\dfrac{\pi}{2}, \dfrac{3\pi}{2}\). \(\cos x = \tfrac12\): \(\dfrac{\pi}{3}, \dfrac{5\pi}{3}\).
  7. If \(\cos x = 0\) then \(\sin^2 x = 1 \ne 0\), so such \(x\) are not solutions, and we may divide by \(\cos^2 x\): \(\tan^2 x = 3\), so \(\tan x = \pm\sqrt3\). \(x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}\).
  8. \(\sin 3x = \sin(2x + x) = \sin 2x\cos x + \cos 2x\sin x = 2\sin x\cos^2 x + (1 - 2\sin^2 x)\sin x\). Replace \(\cos^2 x\) by \(1 - \sin^2 x\): \(2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x = 3\sin x - 4\sin^3 x\). ∎

Common mistake

Splitting a sine like a bracket. \(\sin(a + b)\) is not \(\sin a + \sin b\). Sine is a function, not a number that multiplies. Test with \(a = b = 90°\): \(\sin 180° = 0\), but \(\sin 90° + \sin 90° = 2\).

Missing solutions. A value like \(\sin x = -\tfrac12\) gives two angles in each full turn. Draw the unit circle or the graph and look for all of them. And never divide by a trig function that could be 0.

Where it leads