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11Alexandr11 [23.1K]
3 years ago
9

Cos4theta+cos2theta/ cos4theta-cos2theta= _____

Mathematics
1 answer:
vovangra [49]3 years ago
8 0

\bf \textit{Sum to Product Identities} \\\\ cos(\alpha)+cos(\beta)=2cos\left(\cfrac{\alpha+\beta}{2}\right)cos\left(\cfrac{\alpha-\beta}{2}\right) \\\\\\ cos(\alpha)-cos(\beta)=-2sin\left(\cfrac{\alpha+\beta}{2}\right)sin\left(\cfrac{\alpha-\beta}{2}\right) \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \cfrac{cos(4\theta )+cos(2\theta )}{cos(4\theta )-cos(2\theta )}\implies \cfrac{2cos\left( \frac{4\theta +2\theta }{2} \right)cos\left( \frac{4\theta -2\theta }{2} \right)}{-2sin\left( \frac{4\theta +2\theta }{2} \right)sin\left( \frac{4\theta -2\theta }{2} \right)} \implies \cfrac{cos\left( \frac{6\theta }{2} \right)cos\left( \frac{2\theta }{2} \right)}{-sin\left( \frac{6\theta }{2} \right)sin\left( \frac{2\theta }{2} \right)}

\bf \cfrac{cos(3\theta )cos(\theta )}{-sin(3\theta )sin(\theta )}\implies -\cfrac{cos(3\theta )}{sin(3\theta )}\cdot \cfrac{cos(\theta )}{sin(\theta )}\implies -cot(3\theta )cot(\theta )

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==================================================

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