Is (1-cos(2θ))/(cos(2θ)+1)=tan^2(θ) ?

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

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prove (1-cos(2θ))/(cos(2θ)+1)=tan^2(θ)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{1 - cos ( 2 θ )}{cos ( 2 θ ) + 1}

Rewrite using trig identities

\frac{1 - cos ( 2 θ )}{cos ( 2 θ ) + 1}

Use the Double Angle identity:

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

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

Simplify

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

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

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

1 - 2 sin^{2} (θ) + 1

Group like terms

= - 2 sin^{2} (θ) + 1 + 1

Add the numbers:

1 + 1 = 2

= - 2 sin^{2} (θ) + 2

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

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 ^{2} ( θ ) + 2}

Factor

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

- 2 sin^{2} (θ) + 2

Rewrite as

= - 2 sin^{2} (θ) + 2 \cdot 1

Factor out common term

2

= 2 (- sin^{2} (θ) + 1)

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

Divide the numbers:

\frac{2}{2} = 1

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

Remove parentheses:

(- a) = - a

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

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

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

= \frac{sin ( θ )}{cos ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

\frac{sin ( θ ) tan ( θ )}{cos ( θ )}

= tan (θ) \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

tan (θ) tan (θ)

Simplify

tan (θ) tan (θ) : tan^{2} (θ)

tan (θ) tan (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

tan (θ) tan (θ) = tan^{1 + 1} (θ)

= tan^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= tan^{2} (θ)

tan^{2} (θ)

tan^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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