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Ifa cos θ-b sin e-c, p-/a2 + b2-c2rove that a sin θ + b cos θ |
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Answer» a cosx - b sinx=c squaring both the sides => (acosx - bsinx)² =c² = a²cos²x+b²sin²x-2absinxcosx----------(1)now let's find out the square of (asin x +b cosx)=> (asinx + b cos x)²= a²sin²x+b²cox²x+2absinxcosx --------(2)adding (1) and (2)=> (acosx-bsinx)²+(asinx+bcosx)²=a²cos²x+b²sin²x-2absinxcosx+a²sin²x+b²cox²x+2absinxcosx =>c²+(asinx+bcosx)²=a²cos²x+b²sin²x-2absinxcosx+a²sin²x+b²cox²x+2absinxcosx [using (1)] => c²+(asinx+bcosx)²=b²cos²x+b²sin²x+a²sin²x+a²cox²x => c²+(asinx+bcosx)²= b²(sin²x +cos²x)+ a²(sin²x+cos²x) =>c²+(asinx+bcosx)²= b²+ a² => ( asinx+bcosx)²= b²+ a²-c²=> ( asinx+bcosx) = √(b²+ a²-c²) |
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