Question

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

Answer 1

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

Explanation:

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

Zero

Explanation:

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