Trigonometric Identity Problem

[KingKai]

Homework Statement

Prove the Identity

sinθ/(1+cosθ) = 1-cos(θ)/sinθ

Homework Equations

sinθ/cosθ = tanθ

sin^2θ + cos^2θ = 1

The Attempt at a Solution

sinθ/(1 + cosθ) = LS

cosθtanθ/(1+cosθ) = LS

cosθtanθ/(sin^2θ + cos^2θ + cosθ) = LS

cosθtanθ/(tan^2θcos^2θ + cos^2θ + cosθ) = LS

(cosθ/cosθ) (tanθ/[tan^2θcosθ + cosθ + 1]) = LS

sinθ/(cosθ[tan^2θcosθ + cosθ + 1]) = LS

sinθ/(sin^2θ + cos^2θ +1) = LS

sinθ/2 = LS

 

[SammyS]

Homework Statement

Prove the Identity

sinθ/(1+cosθ) = 1-cos(θ)/sinθ

Homework Equations

sinθ/cosθ = tanθ

sin^2θ + cos^2θ = 1

The Attempt at a Solution

sinθ/(1 + cosθ) = LS

cosθtanθ/(1+cosθ) = LS

cosθtanθ/(sin^2θ + cos^2θ + cosθ) = LS

cosθtanθ/(tan^2θcos^2θ + cos^2θ + cosθ) = LS

(cosθ/cosθ) (tanθ/[tan^2θcosθ + cosθ + 1]) = LS

The above line should be: \displaystyle \frac{\cos(\theta)}{\cos(\theta)}\left(<br /> \frac{\sin(\theta)}{\tan^2(\theta)\cos(\theta) + \cos(\theta) + 1}\right)

sinθ/(cosθ[tan^2θcosθ + cosθ + 1]) = LS

sinθ/(sin^2θ + cos^2θ +1) = LS

sinθ/2 = LS

What is  1 - cos²(θ) ?

How can you make the denominator of LHS or numerator of RHS equal to  1 - cos²(θ) ?

 

[KingKai]

why should sinθ be in the numerator?

I factored out (cosθ/cosθ), where the numerator was initially

cosθtanθ

thus leaving me with tanθ instead of sinθ in the numerator.

 

[SammyS]

why should sinθ be in the numerator?

I factored out (cosθ/cosθ), where the numerator was initially

cosθtanθ

thus leaving me with tanθ instead of sinθ in the numerator.

\displaystyle \cos(\theta)\tan(\theta)=\cos(\theta)\frac{\sin( \theta)}{\cos(\theta)}

\displaystyle =\frac{\cos(\theta)\,\sin(\theta)}{\cos(\theta)}

\displaystyle =\sin(\theta)​

 

[Villyer]

sinθ/(cosθ[tan^2θcosθ + cosθ + 1]) = LS

sinθ/(sin^2θ + cos^2θ +1) = LS

sinθ/2 = LS

In those steps you didn't distribute the cos back to the 1.

 

[ehild]

Homework Statement

Prove the Identity

sinθ/(1+cosθ) = 1-cos(θ)/sinθ

It is not an identity. sinθ/(1+cosθ) = (1-cos(θ))/sinθ is.

ehild

 

[SammyS]

It is not an identity. sinθ/(1+cosθ) = (1-cos(θ))/sinθ is.

ehild

Good point !