Question

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Answer 1

$\displaystyle tgx = \frac{b}{a}\Rightarrow \frac{\sin x}{\cos x} = \frac{b}{a}\Rightarrow a\sin x = b\cos x\Rightarrow \cos x = \frac{a\sin x}{b}$$\displaystyle a\cos2x + b\sin2x = a(1-2\sin^2x) + b\cdot2\sin x\cos x = \\=a-2a\sin^2x+2b\sin x\cdot\frac{a\sin x}{b}=a-2a\sin^2x+2a\sin^2x=a$