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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
If $\cos\alpha+\cos\beta=a$, $\sin\alpha+\sin\beta=b$ and $\theta$ is the arithmetic mean between $\alpha$ and $\beta$, then
$\sin2\theta+\cos2\theta$ is equal to
Solution
$a^2+b^2=(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2$ $=2+2\cos(\alpha-\beta)=4\cos^2\frac{\alpha-\beta}{2}$ $\Rightarrow \cos(\alpha-\beta)=\dfrac{a^2+b^2}{2}-1$ Since $\theta=\dfrac{\alpha+\beta}{2}$, $\sin2\theta+\cos2\theta=\dfrac{a^2-b^2}{a^2+b^2}$ Answer: $\boxed{\dfrac{a^2-b^2}{a^2+b^2}}$Online Test Series, Information About Examination,
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