Question

If $\tan \theta =\frac{a}{b},$ then asin 2θ + b cos 2θ is equal to __________.

Answer 1

$$\begin{gathered} \mathrm{Given}\tan \theta =\frac{a}{b} \\ a\sin 2\theta +b\cos 2\theta \\ =a\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)+b\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right) \\ \left[\mathrm{using}\mathrm{identities}:\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }\mathrm{and}\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right] \\ =\frac{2a\tan \theta +b-b{\tan }^2\theta }{1+{\tan }^2\theta } \\ =\frac{2a\left({\displaystyle \frac{a}{b}}\right)+b-b{\left({\displaystyle \frac{a}{b}}\right)}^2}{1+{\displaystyle \frac{a^2}{b^2}}} \\ =\frac{{\displaystyle \frac{2{\displaystyle a^2}}{b}}+b-{\displaystyle \frac{{\displaystyle a^2}}{b}}}{{\displaystyle \frac{b^2+a^2}{b^2}}} \\ =\frac{{\displaystyle \frac{{\displaystyle a^2}}{b}+b}}{{\displaystyle \frac{\left(b^2+a^2\right)}{b^2}}} \\ =\frac{a^2+b^2}{b}\times \frac{b^2}{a^2+b^2} \\ =b \\ \mathrm{Therefore}, \\ a\sin 2\theta +b\cos 2\theta =b \end{gathered}$$