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

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Answer 1 · 2016-03-27T08:04:04

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

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

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

$cosalphacosbeta=1/2{cos(alpha+beta)+cos(alpha-beta)}$

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

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

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

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

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

Answer 2 · 2016-03-27T14:10:19

Zero

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

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