If tan θ = a/b, Then Find a sin 2θ + b cos 2θ
Question:
\[ \text{If } \tan\theta=\frac{a}{b}, \text{ then find } a\sin2\theta+b\cos2\theta. \]Solution
Given
\[ \tan\theta=\frac{a}{b} \]Using double-angle identities,
\[ \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta} \] \[ =\frac{2(a/b)}{1+a^2/b^2} =\frac{2ab}{a^2+b^2} \]Also,
\[ \cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta} \] \[ =\frac{1-a^2/b^2}{1+a^2/b^2} =\frac{b^2-a^2}{a^2+b^2} \]Therefore,
\[ a\sin2\theta+b\cos2\theta \] \[ = a\left(\frac{2ab}{a^2+b^2}\right) + b\left(\frac{b^2-a^2}{a^2+b^2}\right) \] \[ = \frac{2a^2b+b^3-a^2b}{a^2+b^2} \] \[ = \frac{a^2b+b^3}{a^2+b^2} \] \[ = \frac{b(a^2+b^2)}{a^2+b^2} \] \[ =b \]