Question

How do you find the remaining trigonometric functions of `theta` given `costheta=sqrt2/2` and `theta` terminates in QI?

Answer 1

`"see explanation"`

Explanation:

`"using "sin^2theta+cos^2theta=1`

`rArrsintheta=+-sqrt(1-cos^2theta)`

`"since "theta" is in first quadrant then all ratios are positive"`

`costheta=sqrt2/2`

`•color(white)(x)rArrsintheta=+sqrt(1-((sqrt2)/2)^2)`

`color(white)(rArrsinthetaxxx)=sqrt(1-1/2)=sqrt(1/2)=1/sqrt2=sqrt2/2`

`•color(white)(x)tantheta=sintheta/costheta=(sqrt2/2)/(sqrt2/2)=1`

`•color(white)(x)cottheta=1/tantheta=1`

`•color(white)(x)sectheta=1/costheta=1/(sqrt2/2)=2/sqrt2=sqrt2`

`•color(white)(x)csctheta=1/sintheta=1/(sqrt2/2)=sqrt2`