How do you express $cos (\frac{15 π}{8}) \cdot cos (\frac{5 π}{8})$ $cos( (15 pi)/ 8 ) * cos (( 5 pi) /8 )$ without using products of trigonometric functions?

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How do you express $cos (\frac{15 π}{8}) \cdot cos (\frac{5 π}{8})$ $cos( (15 pi)/ 8 ) * cos (( 5 pi) /8 )$ without using products of trigonometric functions?

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Answer 1 · 2016-04-04T16:08:26

$cos (\frac{15 π}{8}) cos (\frac{5 π}{8}) = \frac{1}{2} cos (\frac{5 π}{2}) + \frac{1}{2} cos (\frac{5 π}{4}) = - \frac{\sqrt{2}}{2}$ $cos((15pi)/8)cos ((5pi)/8)=1/2 cos((5pi)/2)+1/2 cos((5pi)/4)=-sqrt2/2$

Explanation:

$2 cos A cos B = cos (A + B) + cos (A - B)$ $2cos A cos B=cos(A+B)+cos(A-B)$

$cos A cos B = \frac{1}{2} (cos (A + B) + cos (A - B))$ $cosAcos B=1/2 (cos(A+B)+cos(A-B))$

$A = \frac{15 π}{8}, B = \frac{5 π}{8}$ $A=(15pi)/8, B=(5pi)/8$

$\Rightarrow cos (\frac{15 π}{8}) cos (\frac{5 π}{8}) = \frac{1}{2} (cos (\frac{15 π}{8} + \frac{5 π}{8}) + cos (\frac{15 π}{8} - \frac{5 π}{8}))$ $=>cos((15pi)/8)cos ((5pi)/8)=1/2 (cos((15pi)/8+(5pi)/8)+cos((15pi)/8-(5pi)/8))$

$= \frac{1}{2} (cos (\frac{20 π}{8}) + cos (\frac{10 π}{8}))$ $=1/2 (cos((20pi)/8)+cos((10pi)/8))$

$= \frac{1}{2} cos (\frac{5 π}{2}) + \frac{1}{2} cos (\frac{5 π}{4}) = 0 + - \frac{\sqrt{2}}{2} = - \frac{\sqrt{2}}{2}$ $=1/2 cos((5pi)/2)+1/2 cos((5pi)/4) =0+ -sqrt2/2=-sqrt2/2$

$cos (\frac{15 π}{8}) cos (\frac{5 π}{8}) = \frac{1}{2} cos (\frac{5 π}{2}) + \frac{1}{2} cos (\frac{5 π}{4}) = - \frac{\sqrt{2}}{2}$ $cos((15pi)/8)cos ((5pi)/8)=1/2 cos((5pi)/2)+1/2 cos((5pi)/4)=-sqrt2/2$

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How do you express $cos (\frac{15 π}{8}) \cdot cos (\frac{5 π}{8})$ $cos( (15 pi)/ 8 ) * cos (( 5 pi) /8 )$ without using products of trigonometric functions?

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Explanation:

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How do you express cos( (15 pi)/ 8 ) * cos (( 5 pi) /8 ) cos(15π8)⋅cos(5π8)cos( (15 pi)/ 8 ) * cos (( 5 pi) /8 ) without using products of trigonometric functions?

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Explanation:

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How do you express $cos (\frac{15 π}{8}) \cdot cos (\frac{5 π}{8})$ $cos( (15 pi)/ 8 ) * cos (( 5 pi) /8 )$ without using products of trigonometric functions?