Trigonometric Identities

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