How do you express $cos (\frac{3 π}{2}) \cdot cos (\frac{13 π}{8})$ $cos( (3 pi)/ 2 ) * cos (( 13 pi) / 8 )$ without using products of trigonometric functions?

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How do you express $cos (\frac{3 π}{2}) \cdot cos (\frac{13 π}{8})$ $cos( (3 pi)/ 2 ) * cos (( 13 pi) / 8 )$ without using products of trigonometric functions?

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Answer 1 · 2016-03-27T08:04:04

Answer is $0$ $0$ . Please follow the following process.

Explanation:

We know $cos (α + β) = cos α cos β - sin α sin β$ $cos(alpha+beta)=cosalphacosbeta-sinalphasinbeta$
and $cos (α - β) = cos α cos β + sin α sin β$ $cos(alpha-beta)=cosalphacosbeta+sinalphasinbeta$

Adding two we get $cos (α + β) + cos (α - β) = 2 cos α cos β$ $cos(alpha+beta)+cos(alpha-beta)=2cosalphacosbeta$ or

$cos α cos β = \frac{1}{2} {cos (α + β) + cos (α - β)}$ $cosalphacosbeta=1/2{cos(alpha+beta)+cos(alpha-beta)}$

Hence $cos (\frac{3 π}{2}) cos (\frac{13 π}{8}) = \frac{1}{2} {cos (\frac{3 π}{2} + \frac{13 π}{8}) + cos (\frac{3 π}{2} - \frac{13 π}{8})}$ $cos((3pi)/2)cos((13pi)/8)=1/2{cos((3pi)/2+(13pi)/8)+cos((3pi)/2-(13pi)/8)}$

= $\frac{1}{2} {cos (\frac{12 π}{8} + \frac{13 π}{8}) + cos (\frac{12 π}{8} - \frac{13 π}{8})}$ $1/2{cos((12pi)/8+(13pi)/8)+cos((12pi)/8-(13pi)/8)}$

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

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

= $\frac{1}{2} {- cos (\frac{π}{8}) + cos (\frac{π}{8})} = 0$ $1/2{-cos(pi/8)+cos(pi/8)}=0$

Answer 2 · 2016-03-27T14:10:19

Zero

Explanation:

Trig table --> $cos (\frac{3 π}{2})$ $cos ((3pi)/2)$ = 0#, therefor, the product is equal to zero.

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How do you express $cos (\frac{3 π}{2}) \cdot cos (\frac{13 π}{8})$ $cos( (3 pi)/ 2 ) * cos (( 13 pi) / 8 )$ without using products of trigonometric functions?

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Explanation:

Explanation:

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How do you express cos( (3 pi)/ 2 ) * cos (( 13 pi) / 8 ) cos(3π2)⋅cos(13π8)cos( (3 pi)/ 2 ) * cos (( 13 pi) / 8 ) without using products of trigonometric functions?

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Explanation:

Explanation:

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How do you express $cos (\frac{3 π}{2}) \cdot cos (\frac{13 π}{8})$ $cos( (3 pi)/ 2 ) * cos (( 13 pi) / 8 )$ without using products of trigonometric functions?