# Trigonometric Identities Reference

## Sum Identities

$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\cdot\sin\beta$$

$$\cos(\alpha + \beta) = \cos\alpha\cdot\cos\beta - \sin\alpha\cdot\sin\beta$$

$$\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\cdot\tan\beta}$$

$$\tan(\alpha + \beta) = \dfrac{\sin(2\alpha) + \sin(2\beta)}{\cos(2\alpha) + \cos(2\beta)}$$

$$\sin(2\alpha) = 2\sin\alpha\cdot\cos\alpha$$

$$\cos(2\alpha) = \cos^{2}\alpha - \sin^{2}\alpha$$

$$\cos(2\alpha) = 2\cos^{2}\alpha - 1 =$$ $$1 - 2\sin^{2}\alpha$$

$$\tan(2\alpha) = \dfrac{2\tan\alpha}{1 - \tan^{2}\alpha}$$

$$a\sin\theta + b\cos\theta = \sqrt{a^2+b^2}\sin\left(\theta + \arctan\frac{b}{a}\right)$$

## Squares (Pythagorean) Identities

$$\sin^{2}\alpha + \cos^{2}\alpha = 1$$

$$\sec^{2}\alpha = 1 + \tan^{2}\alpha$$

$$\csc^{2}\alpha = 1 + \cot^{2}\alpha$$

## Half-Angle Identities

$$\cos\frac{\alpha}{2} = \sqrt{\frac{1+\cos\alpha}{2}}$$

$$\sin\frac{\alpha}{2} = \sqrt{\frac{1-\cos\alpha}{2}}$$

$$\tan\frac{\alpha}{2} = \sqrt{\frac{1-\cos\alpha}{1-\cos\alpha}}$$

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

$$\tan\frac{\alpha}{2} = \frac{1-\cos\alpha}{\sin\alpha}$$

## Triple-Angle Identities

$$\sin(3\theta) = 3\sin\theta - 4\sin^{3}\theta$$

$$\cos(3\theta) = 4\cos^{3}\theta - 3\cos\theta$$

$$\tan(3\theta) = \frac{3\tan\theta - \tan^{3}\theta}{1-3\tan^{2}\theta}$$

For more multiple-angle identities of sine and cosine, see Cosine Identities.

## Sum to Product Identities

$$\cos m + \cos n = 2\cos\left(\frac{m+n}{2}\right)\cdot \cos\left(\frac{m-n}{2}\right)$$

$$\sin m + \sin n = 2\sin\left(\frac{m+n}{2}\right)\cdot \cos\left(\frac{m-n}{2}\right)$$

## Product to Sum Identities

$$\sin\alpha\cdot\cos\beta = \frac{1}{2}\left[ \sin(\alpha + \beta) + \sin(\alpha - \beta) \right]$$

$$\cos\alpha\cdot\sin\beta = \frac{1}{2}\left[ \sin(\alpha + \beta) - \sin(\beta - \alpha) \right]$$

$$\sin\alpha\cdot\sin\beta = \frac{1}{2}\left[ \cos(\alpha - \beta) - \cos(\alpha + \beta) \right]$$

$$\cos\alpha\cdot\cos\beta = \frac{1}{2}\left[ \cos(\alpha + \beta) + \cos(\alpha - \beta) \right]$$

## Inverse Function Identities

$$\arcsin x \pm \arcsin y = \arcsin\left( x\sqrt{1-y^2} \pm y\sqrt{1-x^2} \right)$$

$$\arccos x \pm \arcsin y = \arccos\left( xy \mp \sqrt{(1-x^2)(1-y^2)} \right)$$

$$\arctan x \pm \arctan y = \arctan\left( \frac{x \pm y}{1 \mp xy}\right)$$

## Composite Identities

$$\sin(\arccos x) = \sqrt{1-x^2}$$

$$\cos(\arcsin x) = \sqrt{1-x^2}$$

$$\sin(\arctan x) = \frac{x}{\sqrt{1+x^2}}$$

$$\cos(\arctan x) = \frac{1}{\sqrt{1+x^2}}$$

$$\tan(\arcsin x) = \frac{x}{\sqrt{1-x^2}}$$

$$\tan(\arccos x) = \frac{\sqrt{1-x^2}}{x}$$

## Interesting Values

$$\cos 20^{\circ}\cdot \cos 40^{\circ} \cdot \cos 80^{\circ} =$$ $$\cos\frac{\pi}{9}\cdot\cos\frac{2\pi}{9}\cdot\cos\frac{4\pi}{9} =$$ $$-\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{4\pi}{7} = \frac{1}{8}$$

$$\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{3\pi}{7} = \frac{1}{8}$$

$$\sin 20^{\circ}\cdot \sin 40^{\circ} \cdot \sin 80^{\circ} =$$ $$\sin\frac{\pi}{9}\cdot\sin\frac{2\pi}{9}\cdot\sin\frac{4\pi}{9} = \frac{\sqrt{3}}{8}$$

$$\sin\frac{\pi}{7}\cdot\sin\frac{2\pi}{7}\cdot\sin\frac{4\pi}{7} =$$ $$\sin\frac{\pi}{7}\cdot\sin\frac{2\pi}{7}\cdot\sin\frac{3\pi}{7} = \frac{\sqrt{7}}{8}$$

$$\tan 50^{\circ}\cdot\tan 60^{\circ}\cdot\tan 70^{\circ} = \tan 80^{\circ}$$

$$\tan 40^{\circ}\cdot\tan 30^{\circ}\cdot\tan 20^{\circ} = \tan 10^{\circ}$$

# Trigonometric Excursions

Problem: If $$\sin x + \cos x = n$$, what is $$\sin^{2} x + \cos^{2} x$$, $$\sin^{3} x + \cos^{3} x$$, and $$\sin^{4} x + \cos^{4} x$$ in terms of n?

Solution: We know that for any x, $$\sin^{2} x + \cos^{2} x = 1$$. We’ll square both sides to find the quantity $$\sin x\cos x$$ however, because this quantity is useful for higher powers.

(i) $$n^2 = (\sin x + \cos x)^2$$

(ii) $$n^2 = \sin^2 x + 2\sin x\cos x + \cos^2 x$$

(iii) $$n^2 = 1 + 2\sin x\cos x$$

(iv) $$\sin x\cos x = \frac{1}{2}(n^2 - 1)$$

Next, we will cube the sum of sine and cosine.

(i) $$n^3 = (\sin x + \cos x)^3$$

(ii) $$n^3 = \sin^3 x + 3\sin^2 x\cos x + 3\sin x\cos^2 x + \cos^3 x$$

(iii) $$n^3 = \sin^3 x + \cos^3 x + 3\sin x\cos x(\sin x + \cos x)$$

(iv) $$\sin^3 x + \cos^3 x = n^3 - 3\cdot\frac{1}{2}(n^2-1)(n)$$  (Substituting the value we found above.)

(v) $$\sin^3 x + \cos^3 x = n^3 - \frac{3}{2}n^3 + \frac{3}{2}n$$

(vi) $$\sin^3 x + \cos^3 x = -\frac{1}{2}n^3 + \frac{3}{2}n$$

Now for the fourth power.

(i) $$n^4 = (\sin x + \cos x)^4$$

(ii) $$n^4 = \sin^4 x + 4\sin^3 x\cos x + 6\sin^2 x\cos^2 x + 4\sin x\cos^3 x + \cos^4 x$$

(iii) $$n^4 = \sin^4 x + \cos^4 x - 2\sin x\cos x (2\sin^2 x + 3\sin x\cos x + 2\cos^2 x)$$

(iv) $$\sin^4 x + \cos^4 x = n^4 - (n^2-1)\left[2 + 3\cdot\frac{1}{2}(n^2-1)\right]$$

(v) $$\sin^4 x + \cos^4 x = -\frac{1}{2}n^4 + n^2+\frac{1}{2}$$