How do you express cos( (3 pi)/ 2 ) * cos (( 5 pi) /4 ) cos(3π2)cos(5π4) without using products of trigonometric functions?

2 Answers
Feb 26, 2016

0

Explanation:

From knowledge of the graph of cosx , we know that

cos((3pi)/2) = 0cos(3π2)=0

and cos((5pi)/4) = -cos(pi/4) = -1/sqrt2 cos(5π4)=cos(π4)=12

rArr cos((3pi)/2) . cos((5pi)/4) = 0.(-1/sqrt2) = 0 cos(3π2).cos(5π4)=0.(12)=0

Feb 26, 2016

It is equivalent to 00.

Explanation:

To express cos(3pi/2)*cos(5pi/4)cos(3π2)cos(5π4), without using trigonometric functions, we should first find value of cos(3pi/2)cos(3π2) and cos(5pi/4)cos(5π4) separately.

cos(3pi/2)cos(3π2) is equal cos((3pi)/2-2pi)cos(3π22π) or cos(-pi/2)cos(π2), which is equal to cos(pi/2)cos(π2). But as latter is equal to zero,

cos((3pi)/2)=0cos(3π2)=0

Although, cos((5pi)/4)=cos(2pi-(5pi)/4)=cos((3pi)/4)=-cos(pi/4)=(-1/sqrt2)cos(5π4)=cos(2π5π4)=cos(3π4)=cos(π4)=(12)

cos(3pi/2)*cos(5pi/4)cos(3π2)cos(5π4) will still be 00, as cos((3pi)/2)=0cos(3π2)=0