How do you express $sin (\frac{π}{4}) \cdot sin (\frac{15 π}{8})$ $sin(pi/ 4 ) * sin( ( 15 pi) / 8 )$ without using products of trigonometric functions?

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How do you express $sin (\frac{π}{4}) \cdot sin (\frac{15 π}{8})$ $sin(pi/ 4 ) * sin( ( 15 pi) / 8 )$ without using products of trigonometric functions?

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Answer 1 · 2016-03-10T10:26:19

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

Explanation:

For this one uses the identity

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

Subtracting upper identity from lower, one gets

$cos (A - B) - cos (A + B) = 2 sin A sin B$ $cos(A-B)-cos(A+B)=2sinAsinB$ or

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

Hence $sin (\frac{5 π}{4}) sin (\frac{15 π}{8})$ $sin((5pi)/4)sin((15pi)/8)$

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

= $\frac{1}{2} (cos (\frac{5 π}{8}) - cos (\frac{25 π}{8}))$ $1/2(cos((5pi)/8)-cos((25pi)/8))$

= $\frac{1}{2} (cos (\frac{5 π}{8}) - cos (3 π + \frac{π}{8}))$ $1/2(cos((5pi)/8)-cos(3pi+pi/8))$

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

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How do you express $sin (\frac{π}{4}) \cdot sin (\frac{15 π}{8})$ $sin(pi/ 4 ) * sin( ( 15 pi) / 8 )$ without using products of trigonometric functions?

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Explanation:

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How do you express $sin (\frac{π}{4}) \cdot sin (\frac{15 π}{8})$ $sin(pi/ 4 ) * sin( ( 15 pi) / 8 )$ without using products of trigonometric functions?