Convert $5 cos (\frac{5 π}{3}) cos (\frac{5 π}{3})$ $5cos((5pi)/3)cos((5pi)/3)$ as a sum of trigonometric ratios?

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Convert $5 cos (\frac{5 π}{3}) cos (\frac{5 π}{3})$ $5cos((5pi)/3)cos((5pi)/3)$ as a sum of trigonometric ratios?

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Answer 1 · 2017-03-29T04:58:04

$5 cos (\frac{5 π}{3}) cos (\frac{5 π}{3}) = \frac{5}{2} (cos (\frac{10 π}{3}) + cos 0)$ $5cos((5pi)/3)cos((5pi)/3)=5/2(cos((10pi)/3)+cos0)$

= $\frac{5}{2} (cos (\frac{10 π}{3}) + 1)$ $5/2(cos((10pi)/3)+1)$

Explanation:

As $cos (A + B) = cos A cos B - sin A sin B$ $cos(A+B)=cosAcosB-sinAsinB$ and

$cos (A - B) = cos A cos B + sin A sin B$ $cos(A-B)=cosAcosB+sinAsinB$

adding the two we get $cos (A + B) + cos (A - B) = 2 cos A cos B$ $cos(A+B)+cos(A-B)=2cosAcosB$

Hence $5 cos (\frac{5 π}{3}) cos (\frac{5 π}{3})$ $5cos((5pi)/3)cos((5pi)/3)$

$= \frac{5}{2} \times 2 cos (\frac{5 π}{3}) cos (\frac{5 π}{3})$ $=5/2xx2cos((5pi)/3)cos((5pi)/3)$

$= \frac{5}{2} \times (cos (\frac{5 π}{3} + \frac{5 π}{3}) + cos (\frac{5 π}{3} - \frac{5 π}{3}))$ $=5/2xx(cos((5pi)/3+(5pi)/3)+cos((5pi)/3-(5pi)/3))$

$= \frac{5}{2} \times (cos (\frac{10 π}{3}) + cos 0)$ $=5/2xx(cos((10pi)/3)+cos0)$ , but $cos 0 = 1$ $cos0=1$

Hence $5 cos (\frac{5 π}{3}) cos (\frac{5 π}{3}) = \frac{5}{2} \times (cos (\frac{10 π}{3}) + 1)$ $5cos((5pi)/3)cos((5pi)/3)=5/2xx(cos((10pi)/3)+1)$

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Convert $5 cos (\frac{5 π}{3}) cos (\frac{5 π}{3})$ $5cos((5pi)/3)cos((5pi)/3)$ as a sum of trigonometric ratios?

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Convert $5 cos (\frac{5 π}{3}) cos (\frac{5 π}{3})$ $5cos((5pi)/3)cos((5pi)/3)$ as a sum of trigonometric ratios?