How do you express $cot^3theta-cos^2theta-tan^2theta$ in terms of non-exponential trigonometric functions?

Question

2821 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-09T15:49:32

$(cos^5(x)-sin^5(x)-cos^4(x)sin^3(x))/(sin^3(x)*cos^2(x))$

We write
$1/tan^3(x)-tan^2(x)-cos^2(x)$

$(1-tan^5(x))/tan^3(x)-cos^2(x)$

$(1-tan^5(x)-tan^3(x)*cos^2(x))/tan^3(x)$

$(1-tan^5(x)-sin^2(x)/cos(x))/tan^3(x)$

$(cos(x)-cos(x)tan^5(x)-sin^3(x))/(cos(x)tan^3(x))$

$(cos^5(x)-sin^5(x)-cos^4(x)sin^3(x))/(sin^3(x)cos^2(x))$

Answer 2 · 2018-08-10T00:31:26

$( cos 3theta + 3 sin theta }/( 3 sin theta - sin 3theta )$
$- 1/2 ( 1 + cos 2theta ) -( 1 - cos 2theta )/( 1 + cos 2theta )$

Use

$cos 3A = 4 cos^3A - 3 cos A and sin 3A = 3 sin A - 4 sin^3A.$

$cot^3theta - cos^2theta - tan^2theta$

$= cos^3theta/sin^3theta - cos^2theta - sin^2theta/cos^2theta$

$= ( cos 3theta + 3 sin theta }/( 3 sin theta - sin 3theta )$

$- 1/2 ( 1 + cos 2theta ) -( 1 - cos 2theta )/( 1 + cos 2theta )$

How do you express cot^3theta-cos^2theta-tan^2theta cot^3theta-cos^2theta-tan^2theta in terms of non-exponential trigonometric functions?