This document contains two methods for proving the identity that tan(2θ)/sin(2θ) = sin(2θ)/cos(2θ). The first method involves manipulating the expressions algebraically, using trigonometric identities like sin(2θ)/cos(2θ) = tan(2θ). The second method works with both sides of the identity separately, again using trigonometric identities and algebraic manipulations to show the two sides are equivalent.

1. Identities
Identities
Identities
Identities
Identities

2. 2.
2. Prove the following
Identity using two
Identity using two
different
different methods:

3. For this question.
You can work
with the more
Method # 1:
complicated side.
In this case it is
the left side.
tan ²θ
sin ²θ
Line of Separation / “Great Wall of China”

4. 1} In step one,
we realized
that we can
multiply by
reciprocal of
1} the bottom
fraction
instead of
dividing.

5. 2} For step
two you
can think
of tanθ as
2} sinθ/cosθ.

6. 3} In step
three we
can
multiply
sinθ by
3} sinθ/cosθ.

7. 4} Here we
can multiply
sin²θ/cosθ
out by
1/cosθcos³θ.
We can also
4} factor out
cos²θ in the
denominator.

8. 5} In this last
step we can
say that
sin²θ/cos²θ is
tan²θ and by
the
Pythagorean
5}
identity (1-
cos²θ) is
sin²θ.

9. QED

10. sinθ tan ²θ
cosθ − cos³θ
sin ²θ
tanθ
 tanθ 
sinθ 
 cosθcos³θ 
 sinθ 
 
sinθ cosθ 
 cosθ − cos³θ 
 
 
sin²θ  1 
 
cosθ  cosθ − cos³θ 
sin²θ
cos²θ − cos 4θ
sin²θ
cos²θ(1 - cos²θ )
tan²θ
sin²θ
QED

11. Working with
both sides
can also
work:
Method # 2:
tan ²θ
sin ²θ
Line of Separation / “Great Wall of China”

12. [Right Side]
tan ²θ 1} Here we
sin ²θ can say that
tan²θ is equal
sin ²θ to
1} cos ²θ sin²θ/cos²θ.
~by saying
sin ²θ that, it can be
1 simplified to
1/cos²θ.
cos ²θ

13. [Left Side]
2} Here we
recognize
cosθ-cos³θ as
cosθ -
(cosθcos²θ)
or cosθ - cosθ
2} (1-sin²θ).
Also, tanθ as
sinθ/cosθ.

14. 3} In this
step we
can
multiply
3} cosθ by
(1-sin²θ).

15. 4} By multiplying
by the reciprocal
of sinθ/cosθ we
can see that
when cosθ gets
multiplied out,
we get cos²θ-
cos²θ+sin²θcosθ.
The two cos²θ’s
4} cancel and
you’re left with
sin²θcosθ.

16. 5} In this last
step we see
that
sin²θcosθ/sinθ
leaves us
sinθcosθ. The
5} sinθ’s reduce
leaving us
1/cosθ.

17. QED

18. sin θ tan ²θ
cos θ − cos ³θ sin ²θ
tan θ sin ²θ
sin θ cos ²θ
cos θ − cos θ (1 − sin ²θ ) sin ²θ
sin θ
1
cos θ
cos ²θ
sin θ
cos θ − cos θ + sin ²θ
sin θ
cos θ
sin θ
sin ²θ cos θ
sin θ
sin θ
sin θ cos θ
1 QED
cos θ

19. Identities
Identities
Identities
Identities
Identities