Question

If sin x/a=cos x/b,prove a sin 2x+b cos2x=b

Answer 1

$$\begin{gathered} \frac{\sin \mathrm{x}}{\mathrm{a}}=\frac{\cos {\displaystyle }{\displaystyle \mathrm{x}}}{\mathrm{b}} \\ \Rightarrow \tan \mathrm{x}=\frac{\mathrm{a}}{\mathrm{b}} \\ \mathrm{Now},\sin \mathrm{x}=\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}};\cos \mathrm{x}=\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}} \\ \mathrm{Now},\mathrm{LHS}=\mathrm{a}\sin 2\mathrm{x}+\mathrm{b}\cos 2\mathrm{x} \\ =\mathrm{a}.2\sin \mathrm{x}.\cos \mathrm{x}+\mathrm{b}\left[2{\cos }^2\mathrm{x}-1\right] \\ =\mathrm{a}\times 2\times \frac{{\displaystyle \mathrm{a}}}{{\displaystyle \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}}\times \frac{{\displaystyle \mathrm{b}}}{{\displaystyle \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}}+\mathrm{b}\left[2\times \frac{{\mathrm{b}}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2}-1\right] \\ =\frac{2{\mathrm{a}}^2\mathrm{b}}{\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}+\frac{2{\mathrm{b}}^3}{\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}-\mathrm{b} \\ =\frac{2{\mathrm{a}}^2\mathrm{b}+2{\mathrm{b}}^3-{\mathrm{a}}^2\mathrm{b}-{\mathrm{b}}^3}{\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)} \\ =\frac{{\mathrm{a}}^2\mathrm{b}+{\mathrm{b}}^3}{\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)} \\ =\frac{\mathrm{b}\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}{\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)} \\ =\mathrm{b} \\ =\mathrm{RHS} \end{gathered}$$