If -1 < p < -1 and sin^3θ - psin^2θ - sinθ + p = 0, then find the value of cosθ.



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Posted on (Answer #1)

Given `-1<p<1` and `sin^3theta-p sin^2theta-sintheta+p=0` find `costheta` :

`sin^3theta-p sin^2theta-sintheta+p=0`

`sin^2theta(sintheta-p)-1(sintheta-p)=0` Factor two terms at a time

`(sintheta-p)(sin^2theta - 1)=0`   Factor out the common factor

If `sin^2theta=+-1` then `costheta=0`

If `sintheta=p` then ` ``costheta=+-sqrt(1-p^2)` where the sign is determined by the quadrant that `theta` is in.

** If `sintheta=p` we can draw a right triangle with acute angle `theta` , side opposite `theta` p, and hypotenuse 1. Then if `costheta=x` we have, from the Pythagorean Theorem, `x^2+p^2=1==>x=+-sqrt(1-p^2)` **


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