Question
$$a \sin B \theta+b \cos B \theta=\sqrt{a^{2}+b^{2}} \cos (B \theta-C)$where $C=\arctan (a / b)$ and $b>0$$
Step 1
We can rewrite this as $\tan C = \frac{a}{b}$. Show more…
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$$a \sin B \theta+b \cos B \theta=\sqrt{a^{2}+b^{2}} \sin (B \theta+C)$ where $C=\arctan (b / a)$ and $a>0$$
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If $\sin \theta+\sin \phi=a$ and $\cos \theta+\cos \phi=b$, then (a) $\cos \frac{\theta-\phi}{2}=\pm \frac{1}{2} \sqrt{a^{2}+b^{2}}$ (b) $\cos \frac{\theta-\phi}{2}=\pm \frac{1}{2} \sqrt{a^{2}-b^{2}}$
Verify the identity. $$\begin{aligned} &a \sin B \theta+b \cos B \theta=\sqrt{a^{2}+b^{2}} \cos (B \theta-C), \text { where }\\ &C=\arctan (a / b) \text { and } b>0. \end{aligned}$$
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