Trigonometric Identities

Pythagorean Identities
Theorem: $$\sin^2 x + \cos^2 x \equiv 1\!$$

Proof:


 * Starting with $$\sin x$$ and $$\cos x$$,

$$\sin x = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$$

$$\cos x = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$$


 * Squaring both sides and adding them together gives:

$$\sin^2 x + \cos^2 x = \frac{\mathrm{opposite}^2 + \mathrm{adjacent}^2}{\mathrm{hypotenuse}^2} \equiv 1$$ by Pythagoras's Theorem

This can be extended to find other trigonometric functions:


 * Starting with $$\sin^2 x + \cos^2 x \equiv 1\!$$, divide every term by $$\cos^2 x \!$$ gives:

$$\tan^2 x + 1 \equiv \sec^2 x \!$$


 * Alternatively, dividing by $$\sin^2 x \!$$ gives:

$$1 + \cot^2 x \equiv \csc^2 x \!$$

Angle Addition and Subtraction Formulas
Theorems: $$\cos(a+b)=\cos a\cos b-\sin a\sin b$$ $$\sin(a+b)=\sin a\cos b+\cos a\sin b$$ $$\cos(a-b)=\cos a\cos b+\sin a\sin b$$ $$\sin(a-b)=\sin a\cos b-\cos a\sin b$$ $$\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$$ $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}$$

Proof:

From Euler's formula we have $$\cos(a+b)+\imath\sin(a+b)=e^{i(a+b)}$$

Using laws of exponents $$\cos(a+b)+\imath\sin(a+b)=e^{\imath a}e^{\imath b}$$

Then again by Euler's formula $$\cos(a+b)+\imath\sin(a+b)=(\cos a+\imath\sin a)(\cos b+\imath\sin b)$$

And distributing $$\cos(a+b)+\imath\sin(a+b)=(\cos a\cos b-\sin a\sin b)+\imath(\sin a\cos b+\cos a\sin b)$$

By equating real and imaginary parts, we have $$\cos(a+b)=\cos a\cos b-\sin a\sin b$$ and $$\sin (a+b)=\sin a\cos b+\cos a\sin b$$

Plugging in -b for b, we get the subtraction formulas because $$\sin(-b)=-\sin b$$ and $$\cos(-b)=\cos b$$

Therefore $$\cos(a-b)=\cos a\cos b+\sin a\sin b$$ and $$\sin(a-b)=\sin a\cos b-\cos a\sin b$$

$$\tan(a+b)=\frac{\sin(a+b)}{\cos(a+b)}$$

$$\tan(a+b)=\frac{\sin a\cos b+\cos a\sin b}{\cos a\cos b-\sin a\sin b}$$

Dividing the numerator and denominator by $$\cos a\cos b$$ we have $$\tan(a+b)=\frac{\frac{\sin a}{\cos a}+\frac{\sin b}{\cos b}}{1+\frac{\sin a\sin b}{\cos a\cos b}}$$

Simplifying, $$\tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$$

Plugging in -b for b, we get the subtraction formula because $$\tan(-b)=-\tan b$$

Therefore $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}$$

Double-Angle Formulas
Theorems:$$\sin(2\theta)=2\sin\theta\cos\theta$$ $$\cos(2\theta)=\cos^2\theta-\sin^2\theta$$ $$\tan(2\theta)=\frac{2\tan\theta}{1-\tan^2\theta}$$

Proof: First, we look for identities to reduce $$\sin(2\theta)$$ and $$\cos(2\theta)$$ Because of the two simple facts from Euler's formula that $$\sin(2\theta) = \mbox {imag}(e^{2i\theta})$$ and $$\cos(2\theta) = \mbox {real}(e^{2i\theta})$$, we need only to look at $$e^{2i\theta}$$ and then extract the appropriate identity.


 * 1) $$e^{2i\theta}$$
 * 2) $$(e^{i\theta})^2$$ (Rules of exponents)
 * 3) $$(i\sin(\theta) + \cos(\theta))^2$$ (Euler's formula)
 * 4) $$i^2\sin^2(\theta) + i\sin(\theta)\cos(\theta) + i\cos(\theta)\sin(\theta) + \cos^2(\theta)$$ (Expanding the polynomial)
 * 5) $$2i\sin(\theta)\cos(\theta) + \cos^2(\theta) - \sin^2(\theta)$$ (Simplify)

From there, by looking at the real and imaginary parts in turn, we find that $$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$ and $$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$$.

Since $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$, we have: $$\tan(2\theta)=\frac{2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}$$ which is equal to $$\frac{2\tan\theta}{1-\tan^2\theta}$$ by dividing the top and bottom by $$\cos^2\theta$$ Q.E.D.

Alternatively, these results follow trivially from the angle addition formulas.

NB: The various other forms of $$\cos(2\theta)$$ that are sometimes presented can be derived by combining this result with the Pythagorean Identities.

Half-Angle Formulas
Theorems: 1. $$\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}}$$ 2. $$\cos\frac{\theta}{2}=\pm\sqrt{\frac{1+\cos\theta}{2}}$$ 3. $$\tan\frac{\theta}{2}=\frac{1-\cos\theta}{\sin\theta}$$ Proofs: 1. $$\cos\theta=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}$$ $$\Longrightarrow\cos\theta=(1-\sin^2\frac{\theta}{2})-\sin^2\frac{\theta}{2}$$ $$\Longrightarrow 2\sin^2\frac{\theta}{2}=1-\cos\theta$$ $$\Longrightarrow\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}}$$ 2. $$\cos\theta=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}$$ $$\Longrightarrow\sin^2\frac{\theta}{2}-(1-\sin^2\frac{\theta}{2})=\cos\theta$$ $$\Longrightarrow 2\sin^2\frac{\theta}{2}=1+\cos\theta$$ $$\Longrightarrow\sin\frac{\theta}{2}=\pm\sqrt{\frac{1+\cos\theta}{2}}$$ 3. $$\tan\frac{\theta}{2}=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}$$ $$=\frac{\pm\sqrt{\frac{1-\cos\theta}{2}}}{\pm\sqrt{\frac{1+\cos\theta}{2}}}$$ $$=\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$ $$=\pm\sqrt{\frac{(1-\cos\theta)(1+\cos\theta)}{(1+\cos\theta)^2}}$$ $$=\pm\sqrt{\frac{1-\cos^2\theta}{(1+\cos\theta)^2}}$$ $$=\pm\sqrt{\frac{\sin^2\theta}{(1+\cos\theta)^2}}$$ $$=\pm\frac{\sin\theta}{1+\cos\theta}$$ Since $$\cos\theta\geq-1\Longrightarrow \cos\theta-1\geq 0$$ and $$\sin\theta$$ and $$\tan\frac{\theta}{2}$$ always have the same sign, we can drop the $$\pm$$, and we have $$\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}$$ By multiplying the top and bottom of the fraction by $$\sqrt{1-\cos\theta}$$, we end up with $$\tan\frac{\theta}{2}=\frac{1-\cos\theta}{\sin\theta}$$

Product-to-Sum Formulas
Theorems: 1. $$\cos\alpha\cos\beta=\frac{\cos(\alpha-\beta)+\cos(\alpha+\beta)}{2}$$ 2. $$\sin\alpha\sin\beta=\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}$$ 3. $$\sin\alpha\cos\beta=\frac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}$$ 4. $$\cos\alpha\sin\beta=\frac{\sin(\alpha+\beta)-\sin(\alpha-\beta)}{2}$$ Proofs: These formulas are proved by expanding the right side using the angle addition formulas 1. $$\frac{\cos(\alpha-\beta)+\cos(\alpha+\beta)}{2}=\frac{(\cos\alpha\cos\beta+\sin\alpha\sin\beta)+(\cos\alpha\cos\beta-\sin\alpha\sin\beta)}{2}$$ $$=\frac{2\cos\alpha\cos\beta}{2}=\cos\alpha\cos\beta$$ 2. $$\frac{\cos(\alpha-\beta)-\cos(\alpha+\beta)}{2}=\frac{(\cos\alpha\cos\beta+\sin\alpha\sin\beta)-(\cos\alpha\cos\beta-\sin\alpha\sin\beta)}{2}$$ $$=\frac{2\sin\alpha\sin\beta}{2}=\sin\alpha\sin\beta$$ 3. $$\frac{\sin(\alpha+\beta)+\sin(\alpha-\beta)}{2}=\frac{(\sin\alpha\cos\beta+\cos\alpha\sin\beta)+(\sin\alpha\cos\beta-\cos\alpha\sin\beta)}{2}$$ $$=\frac{2\sin\alpha\cos\beta}{2}=\sin\alpha\cos\beta$$ 4. $$\frac{\sin(\alpha+\beta)-\sin(\alpha-\beta)}{2}=\frac{(\sin\alpha\cos\beta+\cos\alpha\sin\beta)-(\sin\alpha\cos\beta-\cos\alpha\sin\beta)}{2}$$ $$=\frac{2\cos\alpha\sin\beta}{2}=\cos\alpha\sin\beta$$

Sum-to-Product Formulas
Theorems: 1. $$\sin\alpha+\sin\beta=2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$$ 2. $$\sin\alpha-\sin\beta=2\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)$$ 3. $$\cos\alpha+\cos\beta=2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$$ 4. $$\cos\alpha-\cos\beta=-2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)$$ Proofs: These formulas are proved by expanding the right side using the product-to-sum formulas 1. $$2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)=2\frac{\sin\left(\frac{\alpha+\beta}{2}+\frac{\alpha-\beta}{2}\right)+\sin\left(\frac{\alpha+\beta}{2}-\frac{\alpha-\beta}{2}\right)}{2}$$ $$=2\frac{\sin\frac{2\alpha}{2}+\sin\frac{2\beta}{2}}{2}=\sin\alpha+\sin\beta$$ 2. $$2\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)=2\frac{\sin\left(\frac{\alpha+\beta}{2}+\frac{\alpha-\beta}{2}\right)-\sin\left(\frac{\alpha+\beta}{2}-\frac{\alpha-\beta}{2}\right)}{2}$$ $$=2\frac{\sin\frac{2\alpha}{2}-\sin\frac{2\beta}{2}}{2}=\sin\alpha-\sin\beta$$ 3. $$2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)=2\frac{\cos\left(\frac{\alpha+\beta}{2}-\frac{\alpha-\beta}{2}\right)+\cos\left(\frac{\alpha+\beta}{2}+\frac{\alpha-\beta}{2}\right)}{2}$$ $$=2\frac{\cos\frac{2\beta}{2}+\cos\frac{2\alpha}{2}}{2}=\cos\beta+\cos\alpha=\cos\alpha+\cos\beta$$ 4. $$-2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)=-2\frac{\cos\left(\frac{\alpha+\beta}{2}-\frac{\alpha-\beta}{2}\right)-\cos\left(\frac{\alpha+\beta}{2}+\frac{\alpha-\beta}{2}\right)}{2}$$ $$=-2\frac{\cos\frac{2\beta}{2}-\cos\frac{2\alpha}{2}}{2}=\cos\alpha-\cos\beta$$

Minor Identities
Theorem: $$\tan x(\tan x+\cot x)=sec^2x$$ Proof: $$\tan x(\tan x+\cot x)=\frac{\sin x}{\cos x}(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}) = \frac{\sin^3x+\sin x\cos^2x}{cos^2x\sin x} = \frac{\sin^2x+\cos^2x}{\cos^2x} = \frac{1}{\cos^2x} = \sec^2x$$

Theorem: $$\sec x-\cos x=\sin x\tan x$$ Proof: $$\sec x-\cos x=\frac{1-\cos^2x}{\cos x}=\frac{\sin^2x}{\cos x}=\sin x\tan x$$

Theorem: $$\tan^2x-\sin^2x=\tan^2x\sin^2x$$ Proof:  $$\tan^2x-\sin^2x=\frac{\sin^2x}{\cos^2x}-\frac{\sin^2x\cos^2x}{\cos^2x}=(1-\cos^2x)\frac{\sin^2x}{\cos^2x}=\sin^2x\tan^2x$$

Theorem: $$\sin^4x-\cos^4x=\sin^2x-cos^2x$$ Proof:  $$\sin^4x-\cos^4x=\sin^2x(1-\cos^2x)-\cos^2x(1-\sin^2x)=\sin^2x-\sin^2x\cos^2x-\cos^2x+\sin^2x\cos^2x=\sin^2x-\cos^2x$$

Theorem: $$\csc x-\sin x=\cos x\cot x$$ Proof: $$\csc x-\sin x=\frac{1-\sin^2x}{\sin x}=\frac{\cos^2x}{\sin x}=\cos x\cot x$$

Theorem: $$\sec^2x+\csc^2x=\sec^2x\csc^2x$$ Proof: $$\sec^2x+\csc^2x=\frac{\sin^2x+\cos^2x}{\cos^2x\sin^2x}=\frac{1}{\cos^2x\sin^2x}=\sec^2x\csc^2x$$

Theorem: $$\sec^4x-\tan^4x=\sec^2x+\tan^2x$$ Proof: $$\sec^4x-\tan^4x=\frac{1-\sin^4x}{\cos^4x}=\frac{1-\sin^2x(1-\cos^2x)}{\cos^2x(1-\sin^2x)}= \frac{1-\sin^2x+\sin^2x\cos^2x}{\cos^2x-\sin^2x\cos^2x}$$ $$ = \frac{\cos^2x+\sin^2x\cos^2x}{\cos^2x-\sin^2x\cos^2x}= \frac{1+\sin^2x}{1-\sin^2x}=\frac{1+\sin^2x}{\cos^2x}=\sec^2x+tan^2x$$

Theorem: $$\sin x\tan x+\cos x=\sec x$$ Proof: $$\sin x\tan x+\cos x=\frac{\sin^2x}{\cos x}+\cos x=\frac{\sin^2x+\cos^2x}{\cos x}=\frac{1}{\cos x}=\sec x$$

Theorem: $$\frac{1}{1-\sin x}+\frac{1}{1+\sin x}=2\sec^2x$$ Proof: $$\frac{1}{1-\sin x}+\frac{1}{1+\sin x}=\frac{1+\sin x+1-\sin x}{1-\sin^2x}=\frac{2}{\cos^2x}=2\sec^2x$$

Theorem: $$\frac{1}{1-\sin x}-\frac{1}{1+\sin x}=2\tan x\sec x$$ Proof: $$\frac{1}{1-\sin x}-\frac{1}{1+\sin x}=\frac{1+\sin x-1+\sin x}{1-\sin^2x}=\frac{2\sin x}{\cos^2x}=2\tan x\sec x$$

Theorem: $$\frac{\cos x}{\sec x+\tan x}=1-\sin x$$ Proof: $$\frac{\cos x}{\sec x+\tan x}=\frac{\cos^2x}{1+\sin x}=\frac{1-\sin^2x}{1+\sin x}=\frac{(1+\sin x)(1-\sin x)}{1+\sin x}=1-\sin x$$

Theorem: $$\frac{\sec x+1}{\sec^2x}=\frac{\sin^2x}{\sec x-1}$$ Proof: $$\frac{\sec x+1}{\sec^2x}=\cos^2x(\sec x+1)=\cos x+\cos^2x=\frac{1+\cos x}{\sec x}=\frac{1-\cos^2x}{\frac{1-\cos x}{\cos x}}=\frac{\sin^2x}{\sec x-1}$$

Theorem: $$(\sin x+\cos x)(\tan x+\cot x)=\sec x+\csc x$$ Proof: $$(\sin x+\cos x)(\tan x+\cot x)=(\sin x+\cos x)\frac{\sin^2x+\cos^2x}{\sin x\cos x}=\frac{\sin x+\cos x}{\sin x\cos x}=\sec x+\cos x$$

Theorem: $$\frac{\tan x}{\sec x+1}=\frac{\sec x-1}{\tan x}$$ Proof: $$\frac{\tan x}{\sec x+1}=\frac{\sin x}{1+\cos x}=\frac{\sin^2x}{\sin x(1+\cos x)}=\frac{1-\cos^2x}{\sin x(1+\cos x)}=\frac{1-\cos x}{\sin x}=\frac{\sec x-1}{\tan x}$$

Theorem: $$(a\cos x+b\sin x)^2+(b\cos x-a\sin x)^2=a^2+b^2$$ Proof: $$(a\cos x+b\sin x)^2+(b\cos x-a\sin x)^2=a^2\cos^2x+2ab\cos x\sin x+b^2\sin^2x+b^2\cos^2x-2ab\sin x\cos x+a^2\sin^2x$$ $$=a^2+b^2(\sin^2+\cos^2)=a^2+b^2$$

Theorem: $$\frac{\sin^2x+2\cos x-1}{\sin^2x+3\cos x-3}=\frac{1}{1-\sec x}$$ Proof: $$\frac{\sin^2x+2\cos x-1}{\sin^2x+3\cos x-3}=\frac{1-\cos^2x-2\cos x-1}{1-\cos^2x+3\cos x-3}=\frac{\cos^2x-2\cos x}{\cos^2x-3\cos x +2}=\frac{\cos x(\cos x-2)}{(\cos x-1)(\cos x-2)}=\frac{\cos x}{\cos x-1}=\frac{1}{1-\sec x}$$

Theorem: $$\frac{\cos^2x+3\sin x-1}{\cos^2x+2\sin x+2}=\frac{1}{1+\csc x}$$ Proof: $$\frac{\cos^2x+3\sin x-1}{\cos^2x+2\sin x+2}=\frac{1-\sin^2x+3\sin x-1}{1-\sin^2x+2\sin x+2}=\frac{\sin^2x-3\sin x}{\sin^2x-2\sin x-3}=\frac{\sin x(\sin x-3)}{(\sin x-3)(\sin x+1)}=\frac{\sin x}{\sin x+1}=\frac{1}{1+\csc x}$$