Trigonometry/For Enthusiasts/Less-Used Trig Identities

Triangle Identities

In addition to the Law of Sines, the Law of Cosines, and the Law of Tangents, there are numerous other identities that apply to the three angles A, B, and C of any triangle (where A+B+C=180° and each of A, B, and C is greater than zero). Some of the most notable ones follow:

${\cos }^2(A)+{\cos }^2(B)+{\cos }^2(C)+2\cos (A)\cos (B)\cos (C)=1$

$\sin (A)+\sin (B)+\sin (C)=4\cos \left(\frac{A}{2}\right)\cos \left(\frac{B}{2}\right)\cos \left(\frac{C}{2}\right)$

$\tan (A)+\tan (B)+\tan (C)=\tan (A)\tan (B)\tan (C)$

$\tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)+\tan \left(\frac{B}{2}\right)\tan \left(\frac{C}{2}\right)+\tan \left(\frac{C}{2}\right)\tan \left(\frac{A}{2}\right)=1$

$\cot (A)\cot (B)+\cot (B)\cot (C)+\cot (C)\cot (A)=1$

$\cot \left(\frac{A}{2}\right)\cot \left(\frac{B}{2}\right)\cot \left(\frac{C}{2}\right)=\cot \left(\frac{A}{2}\right)+\cot \left(\frac{B}{2}\right)+\cot \left(\frac{C}{2}\right)$

$\sin (A)\sin (B)\sin (C)=\frac{1}{(\cot (A)+\cot (B))(\cot (B)+\cot (C))(\cot (C)+\cot (A))}$

$\frac{\sin (A)+\sin (B)-\sin (C)}{\sin (A)+\sin (B)+\sin (C)}=\tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)$

Pythagoras

${\sin }^2(x)+{\cos }^2(x)=1$

$1+{\tan }^2(x)={\sec }^2(x)$

$1+{\cot }^2(x)={\csc }^2(x)$

These are all direct consequences of Pythagoras's theorem.

Sum/Difference of angles

$\cos (x\pm y)=\cos (x)\cos (y)\mp \sin (x)\sin (y)$

$\sin (x\pm y)=\sin (x)\cos (y)\pm \sin (y)\cos (x)$

$\tan (x\pm y)=\frac{\tan (x)\pm \tan (y)}{1\mp \tan (x)\tan (y)}$

Product to Sum

$2\sin (x)\sin (y)=\cos (x-y)-\cos (x+y)$

$2\cos (x)\cos (y)=\cos (x-y)+\cos (x+y)$

$2\sin (x)\cos (y)=\sin (x-y)+\sin (x+y)$

Sum and difference to product

$A\sin (x)+B\cos (x)=C\sin (x+y)$ , where $C=\sqrt{A^2+B^2}$ and $y=\pm \arctan \left(\frac{B}{A}\right)$

$\sin (A)\pm \sin (B)=2\sin \left(\frac{A\pm B}{2}\right)\cos \left(\frac{A\mp B}{2}\right)$

$\cos (A)+\cos (B)=2\cos \sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$

$\cos (A)-\cos (B)=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)$

Multiple angle

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

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

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

$\cot (2x)=\frac{\cot (x)-\tan (x)}{2}$

$\csc (2x)=\frac{\cot (x)+\tan (x)}{2}$

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

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

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

$\cos (4x)=8{\cos }^4(x)-8{\cos }^2(x)+1$

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

${\sin }^2(4x)=16[{\sin }^2(x)-5{\sin }^4(x)+8{\sin }^6(x)-4{\sin }^8(x)]$

$\tan (4x)=\frac{4\tan (x)-4{\tan }^3(x)}{1-6{\tan }^2(x)+{\tan }^4(x)}$

$\cos (5x)=16{\cos }^5(x)-20{\cos }^3(x)+5\cos (x)$

$\sin (5x)=16{\sin }^5(x)-20{\sin }^3(x)+5\sin (x)$

$\tan (5x)=\frac{5\tan (x)-10{\tan }^3(x)+{\tan }^5(x)}{1-10{\tan }^2(x)+5{\tan }^4(x)}$

$\cos (6x)=32{\cos }^6(x)-48{\cos }^4(x)+18{\cos }^2(x)-1$

$\cos (7x)=64{\cos }^7(x)-112{\cos }^5(x)+56{\cos }^3(x)-7\cos (x)$

$\sin (7x)=-64{\sin }^7(x)+112{\sin }^5(x)-56{\sin }^3(x)+7\sin (x)$

$\cos (8x)=128{\cos }^8(x)-256{\cos }^6(x)+160{\cos }^4(x)-32{\cos }^2(x)+1$

$\cos (nx)=2\cos (x)\cos ((n-1)x)-\cos ((n-2)x)$

$\sin (nx)=2\cos (x)\sin ((n-1)x)-\sin ((n-2)x)$

These are all direct consequences of the sum/difference formulae

Half angle

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

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

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

${\cos }^2\left(\frac{3x}{2}\right)=2{\cos }^3(x)-\frac{3\cos (x)+1}{2}$

In cases with $\pm$ , the sign of the result must be determined from the value of $\frac{x}{2}$ . These derive from the $\cos (2x)$ formulae.

Power Reduction

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

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

${\tan }^2(x)=\frac{1-\cos (2x)}{1+\cos (2x)}$

Even/Odd

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

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

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

$\csc (-x)=-\csc (x)$

$\sec (-x)=\sec (x)$

$\cot (-x)=-\cot (x)$

Calculus

$\frac{d}{dx}[\sin (x)]=\cos (x)$

$\frac{d}{dx}[\cos (x)]=-\sin (x)$

$\frac{d}{dx}[\tan (x)]={\sec }^2(x)$

$\frac{d}{dx}[\sec (x)]=\sec (x)\tan (x)$

$\frac{d}{dx}[\csc (x)]=-\csc (x)\cot (x)$

$\frac{d}{dx}[\cot (x)]=-{\csc }^2(x)$