Question

If $\tan \theta =\frac{b}{a}$, find the value of $a\cos 2\theta +b\sin 2\theta$

Answer 1

$\tan \theta =\frac{b}{a}\to \sin \theta =\frac{b}{\sqrt{a^2+b^2}},\cos \theta =\frac{a}{\sqrt{a^2+b^2}}$

$\sin 2\theta =2\cos \theta \sin \theta =\frac{2ab}{a^2+b^2}$

$\cos 2\theta =2{\cos }^2\theta -1=\frac{2a^2}{a^2+b^2}-1=\frac{a^2-b^2}{a^2+b^2}$

$=a\cos 2\theta +b\sin 2\theta$

$=\frac{a(a^2-b^2)}{a^2+b^2}+\frac{b\times 2ab}{a^2+b^2}$

$\frac{a^3-a{b}^2+2a{b}^2}{a^2+b^2}=\frac{a^3+a{b}^2}{a^2+b^2}=\frac{a(a^2+b^2)}{(a^2+b^2)}=a$.