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

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

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Answer 1 · 2016-07-04T07:09:33

$cos (\frac{π}{4}) sin (\frac{π}{8}) = \frac{1}{2} sin (\frac{3 π}{8}) - \frac{1}{2} sin (\frac{π}{8})$ $cos(pi/4)sin(pi/8)=1/2sin((3pi)/8)-1/2sin(pi/8)$

Explanation:

As $sin (A + B) = sin A cos B + cos A sin B$ $sin(A+B)=sinAcosB+cosAsinB$ and

$sin (A - B) = sin A cos B - cos A sin B$ $sin(A-B)=sinAcosB-cosAsinB$

and subtracting second equation from first equation we get

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

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

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

= $\frac{sin (\frac{π}{4} + \frac{π}{8})}{2} - \frac{sin (\frac{π}{4} - \frac{π}{8})}{2}$ $sin(pi/4+pi/8)/2-sin(pi/4-pi/8)/2$

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

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

1 Answer

Explanation:

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