Question

cos(theta+phi)cos(theta-phi)=cos^2 theta-sin^2 phi

can anyone can prove this

Answer 1

@Callisto

Answer 2

@iGreen ,@rosedewittbukater ,@frizzank Help me please

Answer 3

@rosedewittbukater

Answer 4

@frizzank

Answer 5

@AkashdeepDeb

Answer 6

\[LHS\]
\[= \cos(\theta + \phi). \cos(\theta - \phi)\]
\[= \frac{1}{2} (2.\cos(\theta + \phi). \cos(\theta - \phi))\]
\[= \frac{1}{2} (\cos(\theta + \phi + \theta - \phi). \cos(\theta + \phi - \theta + \phi))\]
\[= \frac{1}{2} (\cos~2 \theta + \cos~2 \phi)\]
\[= \frac{1}{2} (2 \cos^2 \theta - 1 + (1 - 2 \sin^2 \phi))\]
\[= \frac{1}{2} (2 \cos^2 \theta - 1 + 1 - 2\sin^2 \phi)\]
\[= \frac{1}{2} ~2(\cos^2 \theta - \sin^2 \phi)\]
\[= \cos^2 \theta - \sin^2 \phi\]
\[= RHS\]

\[Q.E.D\]