How do you simplify f(theta)=2cot(theta/2)-csc(theta/4)-cos(theta/4)f(θ)=2cot(θ2)csc(θ4)cos(θ4) to trigonometric functions of a unit thetaθ?

1 Answer
Jul 24, 2018

color(indigo)(f(theta) = (2 * pm sqrt((1 + cos theta) / (1 - cos theta))) - pm 1 / (sqrt(1/2 (1 - pm sqrt(1/2 (1 - cos theta))))) - pm sqrt(1/2 (1 pm sqrt(1/2 (1 + cos theta)))f(θ)=(2±1+cosθ1cosθ)±112(1±12(1cosθ))± 12(1±12(1+cosθ))

Explanation:

![https://www.onlinemathlearning.com/http://double-angle-formula.html](https://useruploads.socratic.org/A0bL1QNkSTKHmUl1Jp77_half%20angle.png)

f(theta) =2 cot (theta/2) - csc (theta/4) - cos (theta/4)f(θ)=2cot(θ2)csc(θ4)cos(θ4)

color(brown)(cot (theta/2) = 1 / tan (theta/2) = pm sqrt((1 + cos theta) / (1 - cos theta))cot(θ2)=1tan(θ2)=±1+cosθ1cosθ

sin (theta/4) = pm sqrt(1 /2 (1 - sin (theta/2)))sin(θ4)=±12(1sin(θ2))

color(brown)(csc (theta/4) = 1 / sin (theta / 4) = pm 1 / (sqrt(1/2 (1 - pm sqrt(1/2 (1 - cos theta)))csc(θ4)=1sin(θ4)=±112(1±12(1cosθ))

cos (theta/4) = pm sqrt(1 /2 (1 + cos (theta/2)))cos(θ4)=±12(1+cos(θ2))

color(brown)(cos (theta / 4) = pm sqrt(1/2 (1 pm sqrt(1/2 (1 + cos theta)))cos(θ4)=± 12(1±12(1+cosθ))

color(indigo)(f(theta) = (2 * pm sqrt((1 + cos theta) / (1 - cos theta))) - pm 1 / (sqrt(1/2 (1 - pm sqrt(1/2 (1 - cos theta))))) - pm sqrt(1/2 (1 pm sqrt(1/2 (1 + cos theta)))f(θ)=(2±1+cosθ1cosθ)±112(1±12(1cosθ))± 12(1±12(1+cosθ))