Trigonometric Identities

Use these fundemental formulas of trigonometry to help solve problems by re-writing expressions in another equivalent form.

\(\sin(x)=\frac{1}{\csc(x)}\)

\(\cos(x)=\frac{1}{\sec(x)}\)

\(\tan(x)=\frac{1}{\cot(x)}\)

\(\sec(x)=\frac{1}{\cos(x)}\)

\(\csc(x)=\frac{1}{\sin(x)}\)

\(\cot(x)=\frac{1}{\tan(x)}\)

\(\tan(x)=\frac{\sin(x)}{\cos(x)}\)

\(\sin(-x)=-\sin(x)\)

\(\cos(-x)=\cos(x)\)

\(\tan(-x)=-\tan(x)\)

\(\sin^2(x)+\cos^2(x)=1\)

\(1+\tan^2(x)=\sec^2(x)\)

\(1+\cot^2(x)=\csc^2(x)\)

\(\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)\)

\(\sin(a-b)=\sin(a)\cos(b)-\cos(a)\sin(b)\)

\(\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\)

\(\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)\)

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

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

\(\sin(x)+\sin(y)=2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})\)

\(\sin(x)-\sin(y)=2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2})\)

\(\cos(x)+\cos(y)=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})\)

\(\cos(x)-\cos(y)=-2\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})\)

\(\sin(2x)=2\sin(x)\cos(x)\)

\(\cos(2x)=\cos^2(x)-\sin^2(x)=1-2\sin^2(x) = 2\cos^2(x)-1\)

\(\sin(\frac{x}{2})=\pm\sqrt{\frac{1-\cos(x)}{2}}\)

\(\cos(\frac{x}{2})=\pm\sqrt{\frac{1+\cos(x)}{2}}\)

\(\tan(\frac{x}{2})=\pm\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=\frac{1-\cos(x)}{\sin(x)}=\frac{\sin(x)}{1+\cos(x)}\)

\(\sin(x)\cos(y)=\frac{\sin(x+y)+\sin(x-y)}{2}\)

\(\cos(x)\cos(y)=\frac{\cos(x+y)+\cos(x-y)}{2}\)

\(\sin(x)\sin(y)=\frac{\cos(x-y)-\cos(x+y)}{2}\)