Question

Note to the viewers:This is not a question but short notes on trigonometric identities that may be of some use.

Answer 1

a-opposite
b-adjacent
c-hypotenuse.
\[\sin \theta=a/c\]
\[\cos \theta=b/c\]
\[\tan \theta=a/b\]
\[\csc \theta=1/\sin \theta=1/a/c=c/a\]
\[\sec \theta=1/\cos \theta=1/b/c=c/b\]
\[\cot \theta=1/\tan \theta=1/a/b=b/a\]

\[\sin ^{2}\theta+ \cos ^{2}\theta=1\]
\[\tan ^{2}\theta+1=\sec ^{2}\theta\]
\[1+\cot^{2}\theta=\csc^{2}\theta\]
\[tan\theta*\cot\theta=1\]

\[sin(A+B)=sinAcosB+sinBcosA\]
\[cos(A+B)=cosAcosB+sinAsinB\]
\[tan(A+B)=(tanA+tanB)/(1-tanA*tanB)\]

\[sin(A-B)=sinAcosB-sinBcosA\]
\[cos(A-B)=cosAcosB-sinAsinB\]
\[tan(A-B)=(tanA-tanB)/(1+tanA*tanB)\]

Answer 2

\[sin(A+B)+sin(A-B)=2sinAcosB\]
\[cos(A+B)+cos(A-B)=2cosAcosB\]

\[sin(A+B)-sin(A-B)=2cosAsinB\]
\[cos(A-B)-cos(A+B)=2sinAsinB\]

\[sin2\theta=2sin\theta\cos\theta\]
\[cos2\theta=cos^2\theta-sin^2\theta\]
\[cos2\theta=1-2sin^2\theta\]
\[cos2\theta=2cos^2\theta-1\]
\[tan2\theta=2tan\theta/(1-tan^2\theta)\]

\[sin3\theta=3sin\theta-4sin^3\theta\]
\[cos3\theta=4cos^3\theta-3cos\theta\]

\[sin(-\theta)=-sin\theta\]
\[cos(-\theta)=cos\theta\]
\[tan(-theta)=-tan\theta\]

Answer 3

\[sin(90-\theta)=cos\theta\]
\[cos(90-\theta)=sin\theta\]
\[tan(90-\theta)=cot\theta\]

\[sin(180-\theta)=sin\theta\]
\[cos(180-\theta)=-cos\theta\]
\[tan(180-\theta)=-tan\theta\]

\[sin(180+\theta)=-sin\theta\]
\[cos(180+\theta)=-cos\theta\]
\[tan180+\theta)=tan\theta\]

\[sin(90+\theta)=cos\theta\]
\[cos(90+\theta)=-sin\theta\]
\[tan(90+\theta)=-cot\theta\]

Answer 4

Nice & thanx @ajprincess :)