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Secondary school › Trigonometry

Trigonometric formulas

Sum and difference, double and half angle, reduction, sum-to-product and product-to-sum formulas. 14 problems with step-by-step solutions.

Solve each problem in your notebook first, then open the solution. In all answers \(k, m \in \mathbb{Z}\).

Problem 1 Sum and difference formulas

Find the exact value.

a) \(\sin 75^\circ\)   b) \(\cos 15^\circ\)   c) \(\tan 75^\circ\)   d) \(\cos 165^\circ\)   e) \(\cot 15^\circ\)   f) \(\sin 105^\circ\)

Show solution
  • a) \(\sin(45^\circ + 30^\circ) = \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} + \frac{\sqrt2}{2}\cdot\frac12 = \frac{\sqrt6 + \sqrt2}{4}\)
  • b) \(\cos(45^\circ - 30^\circ) = \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} + \frac{\sqrt2}{2}\cdot\frac12 = \frac{\sqrt6 + \sqrt2}{4}\)
  • c) \(\tan(45^\circ + 30^\circ) = \frac{1 + \frac{\sqrt3}{3}}{1 - \frac{\sqrt3}{3}} = \frac{3 + \sqrt3}{3 - \sqrt3} = \frac{(3 + \sqrt3)^2}{6} = 2 + \sqrt3\)
  • d) \(\cos 165^\circ = \cos(180^\circ - 15^\circ) = -\cos 15^\circ = -\frac{\sqrt6 + \sqrt2}{4}\)
  • e) \(\cot 15^\circ = \tan 75^\circ = 2 + \sqrt3\)
  • f) \(\sin 105^\circ = \sin(180^\circ - 75^\circ) = \sin 75^\circ = \frac{\sqrt6 + \sqrt2}{4}\)

Problem 2 Sum and difference formulas

\(\sin x = \frac35\) and \(\cos y = \frac{5}{13}\), where \(x\) and \(y\) are acute angles. Find \(\sin(x + y)\), \(\cos(x - y)\) and \(\tan(x + y)\).

Show solution

\(\cos x = \frac45\), \(\sin y = \frac{12}{13}\), \(\tan x = \frac34\), \(\tan y = \frac{12}{5}\).

\(\sin(x + y) = \frac35\cdot\frac{5}{13} + \frac45\cdot\frac{12}{13} = \frac{15 + 48}{65} = \frac{63}{65}\).

\(\cos(x - y) = \frac45\cdot\frac{5}{13} + \frac35\cdot\frac{12}{13} = \frac{20 + 36}{65} = \frac{56}{65}\).

\(\tan(x + y) = \frac{\frac34 + \frac{12}{5}}{1 - \frac34\cdot\frac{12}{5}} = \frac{\frac{63}{20}}{-\frac{16}{20}} = -\frac{63}{16}\).

Problem 3 Double-angle formulas

a) \(\sin x = \frac14\). Find \(\cos 2x\).

b) \(\sin x + \cos x = \frac12\). Find \(\sin 2x\).

c) \(\sin 18^\circ = a\). Write \(\cos 36^\circ\) in terms of \(a\).

Show solution

a) \(\cos 2x = 1 - 2\sin^2 x = 1 - \frac{2}{16} = \frac78\).

b) Square both sides: \(\sin^2 x + 2\sin x\cos x + \cos^2 x = \frac14\), so \(1 + \sin 2x = \frac14\) and \(\sin 2x = -\frac34\).

c) \(\cos 36^\circ = \cos(2\cdot 18^\circ) = 1 - 2\sin^2 18^\circ = 1 - 2a^2\).

Problem 4 Double-angle formulas

a) \(x \in \left(0, \frac{\pi}{2}\right)\) and \(\cos x = \frac13\). Find \(\sin 2x\) and \(\cos 2x\).

b) \(x \in \left(0, \frac{\pi}{2}\right)\) and \(\cos x - \sin x = \frac13\). Find \(\sin 2x\).

c) \(\cos 2x = \frac13\) and \(x\) is acute. Find \(\sin x\), \(\cos x\) and \(\tan x\).

Show solution

a) \(\sin x = \sqrt{1 - \frac19} = \frac{2\sqrt2}{3}\). \(\sin 2x = 2\cdot\frac{2\sqrt2}{3}\cdot\frac13 = \frac{4\sqrt2}{9}\); \(\cos 2x = 2\cos^2 x - 1 = \frac29 - 1 = -\frac79\).

b) Square: \(1 - \sin 2x = \frac19\), so \(\sin 2x = \frac89\).

c) \(\sin^2 x = \frac{1 - \cos 2x}{2} = \frac13\), so \(\sin x = \frac{\sqrt3}{3}\). \(\cos^2 x = \frac{1 + \cos 2x}{2} = \frac23\), so \(\cos x = \frac{\sqrt6}{3}\). \(\tan x = \frac{\sqrt3}{\sqrt6} = \frac{\sqrt2}{2}\).

Problem 5 Half-angle formulas

a) \(x\) is acute and \(\sin x = 0.8\). Find \(\cos\frac{x}{2}\).

b) Write with the half-angle formulas: \(\sin 50^\circ\), \(\cos 70^\circ\), \(\sin 6x\), \(\cos x\).

Show solution

a) \(\cos x = \sqrt{1 - 0.64} = 0.6\). Since \(\frac{x}{2}\) is acute, \(\cos\frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{0.8} = \frac{2}{\sqrt5} = \frac{2\sqrt5}{5}\).

b) \(\sin 50^\circ = 2\sin 25^\circ\cos 25^\circ\); \(\cos 70^\circ = \cos^2 35^\circ - \sin^2 35^\circ\); \(\sin 6x = 2\sin 3x\cos 3x\); \(\cos x = 1 - 2\sin^2\frac{x}{2} = 2\cos^2\frac{x}{2} - 1\).

Problem 6 Using double-angle formulas backwards

Calculate.

a) \(2\sin 15^\circ\cos 15^\circ\)   b) \(\sin 22.5^\circ\cos 22.5^\circ\)   c) \(4\sin\frac{\pi}{12}\cos\frac{\pi}{12}\)

d) \(\cos^2\frac{\pi}{8} - \sin^2\frac{\pi}{8}\)   e) \(2\cos^2 15^\circ - 1\)   f) \(\dfrac{2\tan\frac{\pi}{8}}{1 - \tan^2\frac{\pi}{8}}\)   g) \(\dfrac{\sin 36^\circ}{\sin 12^\circ} - \dfrac{\cos 36^\circ}{\cos 12^\circ}\)

Show solution
  • a) \(\sin 30^\circ = \frac12\)
  • b) \(\frac12\sin 45^\circ = \frac{\sqrt2}{4}\)
  • c) \(2\sin\frac{\pi}{6} = 1\)
  • d) \(\cos\frac{\pi}{4} = \frac{\sqrt2}{2}\)
  • e) \(\cos 30^\circ = \frac{\sqrt3}{2}\)
  • f) \(\tan\frac{\pi}{4} = 1\)
  • g) Common denominator: \(\frac{\sin 36^\circ\cos 12^\circ - \cos 36^\circ\sin 12^\circ}{\sin 12^\circ\cos 12^\circ} = \frac{\sin 24^\circ}{\frac12\sin 24^\circ} = 2\)

Problem 7 Double angle in a quadrant

a) \(\cos x = \frac23\) and \(x\) is in the fourth quadrant. Find \(\sin 2x\).

b) \(\sin a = \frac{5}{13}\). Find \(\cos 2a\).

c) \(\cos\alpha = \frac45\) and \(0 < \alpha < \frac{\pi}{2}\). Find \(\tan\alpha\).

Show solution

a) In the fourth quadrant \(\sin x < 0\): \(\sin x = -\sqrt{1 - \frac49} = -\frac{\sqrt5}{3}\). \(\sin 2x = 2\cdot\left(-\frac{\sqrt5}{3}\right)\cdot\frac23 = -\frac{4\sqrt5}{9}\).

b) \(\cos 2a = 1 - 2\sin^2 a = 1 - \frac{50}{169} = \frac{119}{169}\).

c) \(\sin\alpha = \frac35\), so \(\tan\alpha = \frac34\).

Problem 8 Reduction formulas

Simplify, given that \(0 < x < \frac{\pi}{2}\).

a) \(\cos\left(\frac{\pi}{2} + x\right)\)   b) \(\sin\left(\frac{3\pi}{2} - x\right)\)   c) \(\tan(\pi + x)\)   d) \(\cos(2\pi - x)\)   e) \(\sin(\pi - x)\)   f) \(\cot\left(\frac{3\pi}{2} + x\right)\)

Show solution

With \(\frac{\pi}{2}\) or \(\frac{3\pi}{2}\) the function changes (sin ↔ cos, tan ↔ cot); with \(\pi\) or \(2\pi\) it stays. The sign is the sign of the original function in that quadrant.

  • a) \(\frac{\pi}{2} + x\) is in II, cosine is negative: \(-\sin x\)
  • b) \(\frac{3\pi}{2} - x\) is in III, sine is negative: \(-\cos x\)
  • c) \(\pi + x\) is in III, tangent is positive: \(\tan x\)
  • d) \(2\pi - x\) is in IV, cosine is positive: \(\cos x\)
  • e) \(\pi - x\) is in II, sine is positive: \(\sin x\)
  • f) \(\frac{3\pi}{2} + x\) is in IV, cotangent is negative: \(-\tan x\)

Problem 9 Reduction formulas

Find the exact value.

a) \(\sin 210^\circ\)   b) \(\cos 300^\circ\)   c) \(\tan 225^\circ\)   d) \(\cot 150^\circ\)   e) \(\sin(-135^\circ)\)   f) \(\cos\frac{7\pi}{6}\)

Show solution
  • a) \(\sin(180^\circ + 30^\circ) = -\sin 30^\circ = -\frac12\)
  • b) \(\cos(360^\circ - 60^\circ) = \cos 60^\circ = \frac12\)
  • c) \(\tan(180^\circ + 45^\circ) = \tan 45^\circ = 1\)
  • d) \(\cot(180^\circ - 30^\circ) = -\cot 30^\circ = -\sqrt3\)
  • e) \(\sin(-135^\circ) = -\sin 135^\circ = -\sin 45^\circ = -\frac{\sqrt2}{2}\)
  • f) \(\cos\left(\pi + \frac{\pi}{6}\right) = -\cos\frac{\pi}{6} = -\frac{\sqrt3}{2}\)

Problem 10 Reduction formulas

Simplify.

a) \(\dfrac{\cos(90^\circ - x) + \sin(180^\circ + x) + \cos(360^\circ - x)}{\sin(90^\circ + x) - \cos(180^\circ - x)}\)

b) \(\cos(90^\circ + x) + \sin(270^\circ - x) - \cos(-x) + \sin(180^\circ - x)\)

c) \(\dfrac{\sin 120^\circ\cdot\cos 210^\circ}{\tan 240^\circ\cdot\cot 330^\circ}\)

Show solution

a) Numerator: \(\sin x - \sin x + \cos x = \cos x\). Denominator: \(\cos x + \cos x = 2\cos x\). Result: \(\frac12\).

b) \(-\sin x - \cos x - \cos x + \sin x = -2\cos x\).

c) \(\sin 120^\circ = \frac{\sqrt3}{2}\), \(\cos 210^\circ = -\frac{\sqrt3}{2}\), \(\tan 240^\circ = \sqrt3\), \(\cot 330^\circ = -\sqrt3\). Result: \(\frac{-\frac34}{-3} = \frac14\).

Problem 11 Sum-to-product formulas

Calculate.

a) \(\sin 75^\circ + \sin 15^\circ\)   b) \(\sin 75^\circ - \sin 15^\circ\)   c) \(\cos 75^\circ + \cos 15^\circ\)   d) \(\cos\frac{5\pi}{12} - \cos\frac{\pi}{12}\)

Show solution
  • a) \(2\sin 45^\circ\cos 30^\circ = 2\cdot\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} = \frac{\sqrt6}{2}\)
  • b) \(2\cos 45^\circ\sin 30^\circ = 2\cdot\frac{\sqrt2}{2}\cdot\frac12 = \frac{\sqrt2}{2}\)
  • c) \(2\cos 45^\circ\cos 30^\circ = \frac{\sqrt6}{2}\)
  • d) \(-2\sin\frac{\pi}{4}\sin\frac{\pi}{6} = -2\cdot\frac{\sqrt2}{2}\cdot\frac12 = -\frac{\sqrt2}{2}\)

Problem 12 Sum-to-product formulas

Simplify.

a) \(\dfrac{\sin 2x + \sin 3x + \sin 4x}{\cos 2x + \cos 3x + \cos 4x}\)   b) \(\dfrac{\sin 3x + \sin x}{\cos x - \cos 3x}\)   c) \(\dfrac{\cos x + \cos 5x}{\sin x + \sin 5x}\)   d) \(\dfrac{\sin 7x + \sin 3x}{\cos 7x - \cos 3x}\)

Show solution

a) \(\sin 2x + \sin 4x = 2\sin 3x\cos x\) and \(\cos 2x + \cos 4x = 2\cos 3x\cos x\). The fraction is \(\frac{\sin 3x(2\cos x + 1)}{\cos 3x(2\cos x + 1)} = \tan 3x\).

b) \(\frac{2\sin 2x\cos x}{-2\sin 2x\sin(-x)} = \frac{2\sin 2x\cos x}{2\sin 2x\sin x} = \cot x\).

c) \(\frac{2\cos 3x\cos 2x}{2\sin 3x\cos 2x} = \cot 3x\).

d) \(\frac{2\sin 5x\cos 2x}{-2\sin 5x\sin 2x} = -\cot 2x\).

Problem 13 Product-to-sum formulas

Calculate.

a) \(\cos 75^\circ\cdot\cos 15^\circ\)   b) \(\sin 75^\circ\cdot\sin 15^\circ\)   c) \(\sin 75^\circ\cdot\cos 15^\circ\)   d) \(\sin\frac{3\pi}{8}\cdot\cos\frac{\pi}{8}\)

Show solution
  • a) \(\frac12[\cos 90^\circ + \cos 60^\circ] = \frac12\cdot\frac12 = \frac14\)
  • b) \(-\frac12[\cos 90^\circ - \cos 60^\circ] = \frac14\)
  • c) \(\frac12[\sin 90^\circ + \sin 60^\circ] = \frac12\left(1 + \frac{\sqrt3}{2}\right) = \frac{2 + \sqrt3}{4}\)
  • d) \(\frac12\left[\sin\frac{\pi}{2} + \sin\frac{\pi}{4}\right] = \frac{2 + \sqrt2}{4}\)

Problem 14 Mixed expressions

a) Given \(12x = \pi\), calculate \(\dfrac{\sin 5x + \sin x}{\cos 5x + \cos x}\).

b) Calculate \(\dfrac{1}{\sin 15^\circ} + \dfrac{1}{\cos 15^\circ}\).

c) Calculate \(\dfrac{\sin 80^\circ + 3\cos 10^\circ}{2\sin 80^\circ - \cos 10^\circ}\).

d) Calculate \(\dfrac{\sin 80^\circ + \sin 40^\circ + \sin 60^\circ}{\cos 80^\circ + \cos 40^\circ + \cos 60^\circ}\).

Show solution

a) \(\frac{2\sin 3x\cos 2x}{2\cos 3x\cos 2x} = \tan 3x = \tan\frac{\pi}{4} = 1\).

b) \(\frac{\sin 15^\circ + \cos 15^\circ}{\sin 15^\circ\cos 15^\circ}\). The numerator: \(\sin 15^\circ + \sin 75^\circ = 2\sin 45^\circ\cos 30^\circ = \frac{\sqrt6}{2}\). The denominator: \(\frac12\sin 30^\circ = \frac14\). Result: \(2\sqrt6\).

c) \(\sin 80^\circ = \cos 10^\circ\), so we get \(\frac{4\cos 10^\circ}{\cos 10^\circ} = 4\).

d) \(\sin 80^\circ + \sin 40^\circ = 2\sin 60^\circ\cos 20^\circ\) and \(\cos 80^\circ + \cos 40^\circ = 2\cos 60^\circ\cos 20^\circ\). The fraction is \(\frac{\sin 60^\circ(2\cos 20^\circ + 1)}{\cos 60^\circ(2\cos 20^\circ + 1)} = \tan 60^\circ = \sqrt3\).

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