Is tan^2(x)=((1-cos(2x)))/(1+cos(2x)) ?

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

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Popular Trigonometry >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Simplify

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

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

Add the numbers:

1 + 1 = 2

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

Expand

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

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

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

1 - 1 = 0

= 2 sin^{2} (x)

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

Factor

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

- 2 sin^{2} (x) + 2

Rewrite as

= - 2 sin^{2} (x) + 2 \cdot 1

Factor out common term

2

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

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

Divide the numbers:

\frac{2}{2} = 1

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

Remove parentheses:

(- a) = - a

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

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

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

tan (x) tan (x)

Simplify

tan (x) tan (x) : tan^{2} (x)

tan (x) tan (x)

Apply exponent rule:

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

tan (x) tan (x) = tan^{1 + 1} (x)

= tan^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= tan^{2} (x)

tan^{2} (x)

tan^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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