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To prove:

$$\sin(x)+\sin(3x) = 2\cos(x)\sin(2x)$$

The trigonometric rule $\sin(x+y) = \sin(x)\cos(y)+\cos(x)\sin(y)$ didn't get me that far

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Use :$$\sin A+ \sin B = 2 \sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$$

Therefore :

$$\sin(3x)+\sin(x) =2 \sin \left(\frac{3x+x}{2}\right)\cos \left(\frac{3x-x}{2}\right)= \color{blue}{2\cos(x)\sin(2x)}$$

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Note the following: $\sin(A+B) = \sin A\cos B+\sin B\cos A\tag{1}$

Also note that $\sin(A-B) = \sin A\cos B -\sin B\cos A\tag{2}$

$$(1)+(2) =\sin(A+B)+\sin(A-B) = 2\sin A\cos B$$

Then let $A+B = x,\quad A-B=3x$

We then get $A=3x+B\implies 3x+2B=x\implies B=-x\implies A=2x$

Then substituting this in we arrive at $$\sin x+\sin 3x = 2\sin(2x)\cos(-x)=2\sin(2x)\cos(x)\quad\text{as required}$$

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  • 2
    $\begingroup$ This is also how you can prove the sum-of-sines rule (Jaideep Khare's answer) if you know the sine-of-a-sum rule. $\endgroup$ – David K Apr 27 '17 at 22:01
  • $\begingroup$ @DavidK I like the way you have formatted your sentence. (+ red triangle) $\endgroup$ – Jaideep Khare Apr 28 '17 at 23:42
  • $\begingroup$ @JaideepKhare (+ red triangle)? wtf lol $\endgroup$ – mrnovice Apr 29 '17 at 0:09
  • $\begingroup$ @mrnovice I mean I have upvoted your comment.... Lol. $\endgroup$ – Jaideep Khare Apr 29 '17 at 7:29
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If you use

sin (x + y ) = sinx cosy + cosx siny

sin (x - y ) = sinx cosy - cosx siny

Then

sin (x+ y) + sin (x - y) = 2sinx cosy

Now

substitute x = (P+Q)/2 and y = (P-Q)/2

sin P + sin Q = 2 sin (P+Q/2) cos (P-Q/2)

Use this formula.

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  • $\begingroup$ How is this different from @mrnovice's answer ? $\endgroup$ – Jaideep Khare Apr 27 '17 at 21:48
  • $\begingroup$ I was late.... my apology $\endgroup$ – Umar Apr 27 '17 at 21:51

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