How do you express $cos (\frac{3 π}{2}) \cdot cos (\frac{3 π}{4})$ $cos( (3 pi)/ 2 ) * cos (( 3 pi) /4 )$ without using products of trigonometric functions?

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How do you express $cos (\frac{3 π}{2}) \cdot cos (\frac{3 π}{4})$ $cos( (3 pi)/ 2 ) * cos (( 3 pi) /4 )$ without using products of trigonometric functions?

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Answer 1 · 2016-05-03T09:54:54

$cos (\frac{3 π}{2}) \cdot cos (\frac{3 π}{4}) = 0$ $cos((3pi)/2)*cos((3pi)/4)=0$

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 them

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

or $cos A cos B = \frac{1}{2} \times cos (A + B) + \frac{1}{2} \times cos (A - B)$ $cosAcosB=1/2xxcos(A+B)+1/2xxcos(A-B)$

Hence $cos (\frac{3 π}{2}) \cdot cos (\frac{3 π}{4})$ $cos((3pi)/2)*cos((3pi)/4)$

= $\frac{1}{2} \times cos ((\frac{3 π}{2}) + (\frac{3 π}{4})) + \frac{1}{2} \times cos ((\frac{3 π}{2}) - (\frac{3 π}{4}))$ $1/2xxcos(((3pi)/2)+((3pi)/4))+1/2xxcos(((3pi)/2)-((3pi)/4))$

= $\frac{1}{2} {cos (\frac{9 π}{4}) + cos (\frac{3 π}{4})}$ $1/2{cos((9pi)/4)+cos((3pi)/4)}$

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

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

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How do you express $cos (\frac{3 π}{2}) \cdot cos (\frac{3 π}{4})$ $cos( (3 pi)/ 2 ) * cos (( 3 pi) /4 )$ without using products of trigonometric functions?

1 Answer

Explanation:

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How do you express $cos (\frac{3 π}{2}) \cdot cos (\frac{3 π}{4})$ $cos( (3 pi)/ 2 ) * cos (( 3 pi) /4 )$ without using products of trigonometric functions?