Question

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

Answer 1

0

Explanation:

From knowledge of the graph of cosx , we know that

`cos((3pi)/2) = 0`

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

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

Answer 2

It is equivalent to `0`.

Explanation:

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

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

`cos((3pi)/2)=0`

Although, `cos((5pi)/4)=cos(2pi-(5pi)/4)=cos((3pi)/4)=-cos(pi/4)=(-1/sqrt2)`

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