Trigonometric equation inversion I am trying to invert the following equation to have it with theta as the subject: y=cos theta sin theta −cos theta −sin theta I tried both standard trig as well as trying to reformulate it as a differential equation (albeit I might have chosen an awkward substitution). Nothing seems to stick and I keep on ending up with pretty nasty expressions. Any ideas?

Hallie Stanton

Hallie Stanton

Answered question

2022-11-10

Trigonometric equation inversion
I am trying to invert the following equation to have it with θ as the subject:
y = cos θ sin θ cos θ sin θ
I tried both standard trig as well as trying to reformulate it as a differential equation (albeit I might have chosen an awkward substitution). Nothing seems to stick and I keep on ending up with pretty nasty expressions. Any ideas?

Answer & Explanation

lesinetzgl5

lesinetzgl5

Beginner2022-11-11Added 18 answers

y = cos θ sin θ cos θ sin θ
y cos θ sin θ = cos θ sin θ
y 2 2 y cos θ sin θ + cos 2 θ sin 2 θ = 1 + 2 cos θ sin θ
y 2 y sin ( 2 θ ) + 1 4 sin 2 ( 2 θ ) = 1 + sin ( 2 θ )
This is a quadratic equation for sin ( 2 θ ) that you can solve.
klesstilne1

klesstilne1

Beginner2022-11-12Added 7 answers

Rewrite the equation as ( 1 cos θ ) ( 1 sin θ ) = 1 + y .
Now make the Weierstrass substitution t = tan ( θ / 2 ). It is standard that cos θ = 1 t 2 1 + t 2 and sin θ = 2 t 1 + t 2 . So our equation becomes
2 t 2 1 + t 2 ( 1 t ) 2 1 + t 2 = 1 + y .
Take the square root, and clear denominators. We get
2 t ( 1 t ) = 1 + y ( 1 + t 2 ) .
This is a quadratic in t.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?