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I need help writing the expression $5(\sin 2x- \cos 2x)$ in terms of sine only. Using the double angle identity formula, I was able to get it to $5(\sin 2x - 1 + 2\sin^2x)$, but not sure how to move forward. I know the answer is $$5\sqrt{2}\sin\left(2x+ \frac{7\pi}{4}\right)$$ but I'm not sure how to get there. Thanks.

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Given $a\sin(u)+b\cos(v)$, factor out whatever you need to to make the coefficients of $\sin$ and $\cos$ have sum of squares equal to $1$. Then a sum of angles formula applies. In this case,

$$5\sqrt{2}\left(\sin(2x)\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\cos(2x)\right) $$

$$=5\sqrt{2}\big[\sin(2x)\cos(-\pi/4)+\sin(-\pi/4)\cos(2x)\big] $$

$$ =5\sqrt{2}\sin(2x-\pi/4).$$

(Of course, $-\pi/4$ works the same as $2\pi-\pi/4=7\pi/4$.)

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Note that $\cos(7\pi/4) = 1/\sqrt{2}$ and $\sin(7\pi/4) = -1\sqrt{2}$ so your expression $$=5\sqrt{2}(\cos(7\pi/4)\sin 2x +\sin(7\pi/4)\cos(2x) = 5\sqrt{2}\sin(2x+7\pi/4).$$

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  • $\begingroup$ Thank you for helping. I have been able to work this problem out to 5sqrt2(1/sqrt2sin2x - 1/sqrt2cos2x), but not sure how to derive result to be in quadrant 4 with sine positive and cosine negative (this would be quadrant 2 and 3pi/4, not 7pi/4). What am I missing? $\endgroup$ – Michael Servilla Oct 19 '16 at 12:34
  • $\begingroup$ I think I see this now, using the example below from artic tern. Thanks! $\endgroup$ – Michael Servilla Oct 19 '16 at 12:40
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Write

$5(\sin2x-\cos2x) = a \sin(bx+c)$

Then

$5\sin2x-5\cos2x = a \sin(bx)\cos(c)+a \sin(c)\cos(bx)$

Choose $b=2$, $a\cos(c)=5$ and $a\sin(c)=-5$. So

$a^2\cos^2(c)+a^2\sin^2(c) = 25+25 = 50 \Rightarrow a^2=50 \Rightarrow a = \pm 5\sqrt{2}$

Choose $a = 5\sqrt{2}$

$\Rightarrow\cos(c) = \dfrac{5}{5\sqrt{2}}=\dfrac{\sqrt{2}}{2}$ and

$\sin(c) = \dfrac{-5}{5\sqrt{2}}=-\dfrac{\sqrt{2}}{2}$

Hence $c = \dfrac{7\pi}{4}$

Then

$5(\sin2x-\cos2x) = 5\sqrt{2} \sin\left(2x+\dfrac{7\pi}{4}\right)$

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