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Express in the form $A\sin(x+c)$
a) $\sin x+\sqrt3\cos x$; b) $\sin x-\cos x$

sol: a) $A=\sqrt{1+3}=2$, $\tan c=\frac{\sqrt 3}1$, $c=\frac\pi3$. So $\sin x+\sqrt3\cos x=2\sin(x+\frac\pi3)$
b) $\sqrt 2\sin(x-\frac\pi4)$

Can someone please explain the method used in the provided solution above? (I'm not familiar with this way of solving whatsoever.)

Thanks in advance =]

  • 1
    There are a few spiritual duplicates of this question. The key is to normalize and use the trigonometric addition formulas. Try and work backwards: expand out $A\sin(x+c)$, and then figure out $A$ and $c$ (do use the formula $\cos^2+\sin^2=1$ to find out $A$).2012-09-10
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    Thanks, I tried expanding the desired form and it all worked out nicely. Though I didn't understand where the identity $cos^2\alpha+sin^2\alpha=1$ comes to use.2012-09-10
  • 0
    If you know $A\cos(c)$ and $A\sin(c)$, then you can find $A$ via $A^2=(A\cos c)^2+(A\sin c)^2$.2012-09-10
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    Oh, I just compared $Acos(c)$ with its value after finding the angle c.2012-09-10
  • 0
    A similar question was posted here http://math.stackexchange.com/questions/877499/express-sinx-sqrt3-cosx-in-the-form-a-sinx-a However, it was now autodeleted.2014-08-05

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