Question

How do you find the other trigonometric functions given `sin(pi/2-x)=sqrt2/2, sinx=-sqrt2/2`?

Answer 1

`"see explanation"`

Explanation:

`"given "sin(pi/2-x)=sqrt2/2`

`rArrcosx=sqrt2/2" and "sinx=-sqrt2/2`

`rArrx" is in the fourth quadrant"`

`•color(white)(x)secx=1/cosx=1/(sqrt2/2)=2/sqrt2=sqrt2`

`•color(white)(x)cscx=1/sinx=1/(-sqrt2/2)=-2/sqrt2=-sqrt2`

`•color(white)(x)tanx=sinx/cosx=(-sqrt2/2)/(sqrt2/2)=-1`

`•color(white)(x)cotx=1/tanx=-1`