Verifying Trigonometric Identities

Question

verify

sin²θ/(1+cosθ) = (tanθ - sinθ)/tanθ

Answer 1

Hi Bryan!

Sometimes you can start with one side and

arrive at the other. In this case, you actually

need to simplify both sides.

I'll start with the right side first:

tanθ - sinθ sinθ

---------------- tanθ = -----

tanθ cosθ

sinθ sinθ

----- - -----

cosθ 1

----------------------

sinθ

------

cosθ

sinθ sinθ cosθ Get common

----- - ----- * ------ denominator

cosθ 1 cosθ of cosθ

---------------------------------

sinθ

------

cosθ

sinθ - sinθcosθ

----------------------

cosθ

--------------------------

sinθ

-----

cosθ

Multiply and cosθ cancels out

sinθ - sinθcosθ cosθ

----------------------- * -------

cosθ sinθ

Factor out sinθ. sinθ cancels out

sinθ (1 - cosθ)

-------------------

sinθ

(1- cosθ) or 1- cosθ

Now, simplify the left side

sin²θ

------------------ use pythagorean identity

1 + cosθ sin²θ + cos²θ = 1

sin²θ = 1- cos²θ

1-cos²θ

--------------- 1-cos²θ =(1-cosθ)(1+cosθ)

1 + cosθ difference of two squares

(1-cosθ)(1+cosθ)

------------------------ (1+cosθ) cancels

1 + cosθ

1-cosθ

So, both sides simplify to 1-cosθ.

Note** Raymond's solution is much shorter.

Here is another way to do it.

Answer 2

sin^2T/(1+cosT)

=(1-cos^2T)/(1+cosT) Pythagorean Theorem

=(1+cosT)(1-cosT)/(1+cosT) factor difference of squares=product of sum and difference

=1-cosT cancel common factor 1+cosT

=tanT/tanT -cosTtanT/tanT multiply by 1= tanT/tanT

=(tanT-cosTtanT)/tanT combine over common denominator

=(tanT-cosTsinT/cosT)/tanT replace tanT with sinT/cosT

=(tanT-sinT)/tanT cancel common factor cosT

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