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How to show the equation $4 \sin(x) +3 \cos (x)$ as $r \sin (x-\alpha)$

I tried to solve this using $a \cos(x)+b \sin(x) = c$. But there aren't any to use as $c$

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  • $\begingroup$ You want to solve $a \cos(x)+b \sin(x) = r \sin (x-\alpha)$, don't you ? So your $c$ is $r \sin (x-\alpha)$, simply. $\endgroup$
    – user65203
    Feb 17, 2015 at 8:08

4 Answers 4

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Because $$\left(\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2=1,$$ taking $\alpha=\arccos(4/5)$ gives that $$\cos\alpha=4/5,\quad\sin\alpha=3/5.$$ Hence, $$4\sin x +3\cos x=5(\cos\alpha\sin x+\sin\alpha\cos x)=5\sin(x+\alpha).$$

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The angle-addition formula $$ \sin(x-\alpha) = \sin(x)\cos(\alpha) - \cos(x)\sin(\alpha) $$ will help you set up a system of equations for $r$ and $\alpha$.

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By your method,
Since $$\sin(A+B)=\sin A\cos B +\sin B\cos A$$ Therefore,
$$r\cos\alpha=4$$ and $$r\sin\alpha=3$$ Dividing them $$\tan\alpha=\frac{3}{4}$$ or $$\alpha=\arctan\frac{3}{4}\approx 37^{\circ}$$ $$\implies \sin\alpha=\frac{3}{5}$$ We get $$r\left(\frac{3}{5}\right)=3$$ $$r=5$$
We finally get, $$5\sin\left(x+\arctan\frac{3}{4}\right)$$ $$\approx 5\sin(x+37^{\circ})$$

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Hint: you should normalize this $$a \cdot sin(x) + b \cdot cos(x) = \sqrt{(a+b)} \cdot \bigg(\frac{a\cdot \sin(x) }{\sqrt{(a+b)}} + \frac{b \cdot \cos(x) }{\sqrt{(a+b)}}\bigg)$$ And than use formula from the post below.

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