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

2 Answers
Mar 27, 2016

Answer is 00. Please follow the following process.

Explanation:

We know cos(alpha+beta)=cosalphacosbeta-sinalphasinbetacos(α+β)=cosαcosβsinαsinβ
and cos(alpha-beta)=cosalphacosbeta+sinalphasinbetacos(αβ)=cosαcosβ+sinαsinβ

Adding two we get cos(alpha+beta)+cos(alpha-beta)=2cosalphacosbetacos(α+β)+cos(αβ)=2cosαcosβ or

cosalphacosbeta=1/2{cos(alpha+beta)+cos(alpha-beta)}cosαcosβ=12{cos(α+β)+cos(αβ)}

Hence cos((3pi)/2)cos((13pi)/8)=1/2{cos((3pi)/2+(13pi)/8)+cos((3pi)/2-(13pi)/8)}cos(3π2)cos(13π8)=12{cos(3π2+13π8)+cos(3π213π8)}

= 1/2{cos((12pi)/8+(13pi)/8)+cos((12pi)/8-(13pi)/8)}12{cos(12π8+13π8)+cos(12π813π8)}

= 1/2{cos((25pi)/8)+cos(-pi/8)}12{cos(25π8)+cos(π8)}

= 1/2{cos(2pi+pi+pi/8)+cos(pi/8)}12{cos(2π+π+π8)+cos(π8)}

= 1/2{-cos(pi/8)+cos(pi/8)}=012{cos(π8)+cos(π8)}=0

Mar 27, 2016

Zero

Explanation:

Trig table --> cos ((3pi)/2)cos(3π2) = 0#, therefor, the product is equal to zero.