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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Using the fact that $\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$and the differentiation, obtain the sum formula for cosines.
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Answer

$\sin(\text{A}+\text{B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}$
Differentiating both sides with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}[\sin(\text{A}+\text{B})]=\frac{\text{d}}{\text{dx}}(\sin\text{A}\cos\text{B})+\frac{\text{d}}{\text{dx}}(\cos\text{A}\sin\text{B})$
$\Rightarrow\ \cos(\text{A}+\text{B}).\frac{\text{d}}{\text{dx}}(\text{A}+\text{B})$
$=\cos\text{B}.\frac{\text{d}}{\text{dx}}(\sin\text{A})+\sin\text{A}.\frac{\text{d}}{\text{dx}}(\cos\text{B})+\sin\text{B}.\frac{\text{d}}{\text{dx}}(\cos\text{A})+\cos\text{A}.\frac{\text{d}}{\text{dx}}(\sin\text{B})$
$\Rightarrow\ \cos(\text{A}+\text{B}).\frac{\text{d}}{\text{dx}}(\text{A}+\text{B})$
$=\cos\text{B}.\cos\text{A}\frac{\text{dA}}{\text{dx}}+\sin\text{A}(-\sin\text{B})\frac{\text{dB}}{\text{dx}}+\sin\text{B}(-\sin\text{A})\frac{\text{dA}}{\text{dx}}+\cos\text{A}\cos\text{B}\frac{\text{dB}}{\text{dx}}$
$\Rightarrow\ \cos(\text{A}+\text{B}).\Big[\frac{\text{dA}}{\text{dx}}+\frac{\text{dB}}{\text{dx}}\Big]$
$=(\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}).\Big[\frac{\text{dA}}{\text{dx}}+\frac{\text{dB}}{\text{dx}}\Big]$
$\therefore\ \cos(\text{A}+\text{B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$

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