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Using the fact that \(\sin (A + B) = \sin A \cos B + \cos A \sin B\) and the differentiation, obtain the sum formula for cosines.

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  • $\cos (A + B) = \cos A \cos B - \sin A \sin B$
Step 1:
$\sin (A + B) = \sin A \cos B + \cos A \sin B$------(1)
Consider A and B as functions of $t$ and differentiating both sides of (1) with respect to $t$ we have
$\cos(A+B)\big(\large\frac{dA}{dt}+\frac{dB}{dt}\big)=$$\begin{bmatrix}\sin A(-\sin B)\large\frac{dB}{dt}+\normalsize\cos B\cos A\large\frac{dA}{dt}\end{bmatrix}$+$\begin{bmatrix}\cos A\cos B\large\frac{dB}{dt}\normalsize+\sin B(-\sin A)\large\frac{dA}{dt}\end{bmatrix}$
$\Rightarrow \cos(A+B)\big(\large\frac{dA}{dt}+\frac{dB}{dt}\big)$
Step 2:
We know that $\cos (A + B) = \cos A \cos B - \sin A \sin B$
$\Rightarrow (\cos A \cos B - \sin A \sin B)\big(\large\frac{dA}{dt}+\frac{dB}{dt}\big)$
$\Rightarrow \cos(A+B)\big(\large\frac{dA}{dt}+\frac{dB}{dt}\big)$=$(\cos A \cos B - \sin A \sin B)\big(\large\frac{dA}{dt}+\frac{dB}{dt}\big)$
$\Rightarrow\cos (A + B) = \cos A \cos B - \sin A \sin B$
answered May 15, 2013 by sreemathi.v
 

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