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The question is to show that $A\sin(x + B)$ can be written as $a\sin x + b\cos x$ for suitable a and b.

Also, could somebody please show me how $f(x)=A\sin(x+B)$ satisfies $f + f ''=0$?

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up vote 2 down vote accepted

If $$ f(x) = A\sin(x+B) $$ then $$ f'(x) = A\cos(x+B)\cdot\frac{d}{dx}(x+B) = A\cos(x+B)\cdot1, $$ and $$ f''(x) = -A\sin(x+B)\cdot\frac{d}{dx}(x+B) = -A\sin(x+B). $$ So $$ f''(x)+f(x) = -A\sin(x+B)+A\sin(x+B) = 0. $$

For the initial question, the standard trigonometric identity $$ \sin(x+B) = \sin x\cos B+ \cos x\sin B $$ is most of what you need to know. Then you have $$ A\sin(x+B) = A\Big( \sin x\cos B+ \cos x\sin B \Big) $$ $$ = \Big(A\cos B\Big) \sin x + \Big( A\sin B\Big) \cos x $$ $$ =a\sin x+b\cos x. $$

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Thank you so much. Does this would mean that a=f '(0)=AcosB and b=f(0)=AsinB? – Dick Feb 2 '13 at 19:58
    
Yes. ${{{{{{{{{{}}}}}}}}}}$ – Michael Hardy Feb 2 '13 at 20:09

Given $f(x)=\sin(x+B)$ the chain rule gives $f'(x)=(x+B)'\cos(x+B)=\cos (x+B)$. Then another derivative gives $f''(x)=-\sin (x+B)$

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