Question

Question #110e4

Answer 1

`cos(theta)=-sqrt(1-k^2)`
`tan(theta)=-k/sqrt(1-k^2)=-(ksqrt(1-k^2))/(1-k^2)`
`sin(theta+pi)=-k`

Explanation:

It is given that `sin(theta)=k` and that `theta` is an obtuse angle (`pi/2< theta< pi`). Thus, angle `theta` is in the second quadrant.

We can find `cos(theta)` by using the property `sin^2(theta)+cos^2(theta)=1`. Then, `cos(theta)=+-sqrt(1-sin^2(theta))=+-sqrt(1-k^2)`. Since `theta` is in the second quadrant and the cosine function returns negative values in the second quadrant, `cos(theta)=-sqrt(1-k^2)`.

We can find `tan(theta)` by using the property `tan(theta)=sin(theta)/cos(theta)=k/-sqrt(1-k^2)=-k/sqrt(1-k^2)`. This can also be rewritten as `-(ksqrt(1-k^2))/(1-k^2)`.

To find `sin(theta+pi)`, use the unit circle. `pi` is half a circle, and adding it to an angle negates both its `x` and `y` coordinate. `sin(theta)` is the value of the `y` coordinate; therefore `sin(theta+pi)=-sin(theta)=-k`.