Learn the fundamental trig identities
Identities are equations that are true for all values of the variable.
sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1sin2θ+cos2θ=1
Derived from this: tan2θ+1=sec2θ\tan^2\theta + 1 = \sec^2\thetatan2θ+1=sec2θ 1+cot2θ=csc2θ1 + \cot^2\theta = \csc^2\theta1+cot2θ=csc2θ
tanθ=sinθcosθ\tan\theta = \frac{\sin\theta}{\cos\theta}tanθ=cosθsinθ cotθ=cosθsinθ\cot\theta = \frac{\cos\theta}{\sin\theta}cotθ=sinθcosθ
cscθ=1sinθ\csc\theta = \frac{1}{\sin\theta}cscθ=sinθ1 secθ=1cosθ\sec\theta = \frac{1}{\cos\theta}secθ=cosθ1 cotθ=1tanθ\cot\theta = \frac{1}{\tan\theta}cotθ=tanθ1
Cosine is even: cos(−θ)=cosθ\cos(-\theta) = \cos\thetacos(−θ)=cosθ Sine is odd: sin(−θ)=−sinθ\sin(-\theta) = -\sin\thetasin(−θ)=−sinθ Tangent is odd: tan(−θ)=−tanθ\tan(-\theta) = -\tan\thetatan(−θ)=−tanθ
sin(A+B)=sinAcosB+cosAsinB\sin(A + B) = \sin A \cos B + \cos A \sin Bsin(A+B)=sinAcosB+cosAsinB sin(A−B)=sinAcosB−cosAsinB\sin(A - B) = \sin A \cos B - \cos A \sin Bsin(A−B)=sinAcosB−cosAsinB
cos(A+B)=cosAcosB−sinAsinB\cos(A + B) = \cos A \cos B - \sin A \sin Bcos(A+B)=cosAcosB−sinAsinB cos(A−B)=cosAcosB+sinAsinB\cos(A - B) = \cos A \cos B + \sin A \sin Bcos(A−B)=cosAcosB+sinAsinB
tan(A+B)=tanA+tanB1−tanAtanB\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}tan(A+B)=1−tanAtanBtanA+tanB
sin(2θ)=2sinθcosθ\sin(2\theta) = 2\sin\theta\cos\thetasin(2θ)=2sinθcosθ
cos(2θ)=cos2θ−sin2θ\cos(2\theta) = \cos^2\theta - \sin^2\thetacos(2θ)=cos2θ−sin2θ =2cos2θ−1= 2\cos^2\theta - 1=2cos2θ−1 =1−2sin2θ= 1 - 2\sin^2\theta=1−2sin2θ
tan(2θ)=2tanθ1−tan2θ\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}tan(2θ)=1−tan2θ2tanθ
Example: Simplify sin2θ1+cosθ\frac{\sin^2\theta}{1 + \cos\theta}1+cosθsin2θ
=1−cos2θ1+cosθ= \frac{1 - \cos^2\theta}{1 + \cos\theta}=1+cosθ1−cos2θ =(1−cosθ)(1+cosθ)1+cosθ= \frac{(1 - \cos\theta)(1 + \cos\theta)}{1 + \cos\theta}=1+cosθ(1−cosθ)(1+cosθ) =1−cosθ= 1 - \cos\theta=1−cosθ