Is tan(θ)=(1-cos(2θ))/(sin(2θ)) ?

The answer to whether tan(θ)=(1-cos(2θ))/(sin(2θ)) is True

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prove tan(θ)=(1-cos(2θ))/(sin(2θ))

prove

tan (θ) = \frac{1 - cos ( 2 θ )}{sin ( 2 θ )}

True

Solution steps

tan (θ) = \frac{1 - cos ( 2 θ )}{sin ( 2 θ )}

Manipulating right side

\frac{1 - cos ( 2 θ )}{sin ( 2 θ )}

Rewrite using trig identities

\frac{1 - cos ( 2 θ )}{sin ( 2 θ )}

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= \frac{1 - cos ( 2 θ )}{2 sin ( θ ) cos ( θ )}

Use the Double Angle identity:

cos (2 x) = 1 - 2 sin^{2} (x)

= \frac{1 - ( 1 - 2 sin ^{2} ( θ ) )}{2 sin ( θ ) cos ( θ )}

Simplify

\frac{1 - ( 1 - 2 sin ^{2} ( θ ) )}{2 sin ( θ ) cos ( θ )} : \frac{sin ( θ )}{cos ( θ )}

\frac{1 - ( 1 - 2 sin ^{2} ( θ ) )}{2 sin ( θ ) cos ( θ )}

Expand

1 - (1 - 2 sin^{2} (θ)) : 2 sin^{2} (θ)

1 - (1 - 2 sin^{2} (θ))

- (1 - 2 sin^{2} (θ)) : - 1 + 2 sin^{2} (θ)

- (1 - 2 sin^{2} (θ))

Distribute parentheses

= - (1) - (- 2 sin^{2} (θ))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + 2 sin^{2} (θ)

= 1 - 1 + 2 sin^{2} (θ)

1 - 1 = 0

= 2 sin^{2} (θ)

= \frac{2 sin ^{2} ( θ )}{2 sin ( θ ) cos ( θ )}

Divide the numbers:

\frac{2}{2} = 1

= \frac{sin ^{2} ( θ )}{sin ( θ ) cos ( θ )}

Cancel the common factor:

sin (θ)

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

= tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

Is tan(θ)=(1-cos(2θ))/(sin(2θ)) ?
The answer to whether tan(θ)=(1-cos(2θ))/(sin(2θ)) is True