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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{- 1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}

Simplify

\frac{- 1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}} : \frac{- cos ^{2} ( θ ) + sin ^{2} ( θ )}{cos ^{2} ( θ ) + sin ^{2} ( θ )}

\frac{- 1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{- 1 + ( \frac{s i n ( θ )}{c o s ( θ )} ) ^{2}}{1 + \frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{- 1 + \frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}{1 + \frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

= \frac{- 1 + \frac{s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}{\frac{c o s ^{2} ( θ ) + s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

= \frac{\frac{- c o s ^{2} ( θ ) + s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}{\frac{c o s ^{2} ( θ ) + s i n ^{2} ( θ )}{c o s ^{2} ( θ )}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

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

Cancel the common factor:

cos^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Apply rule

\frac{a}{1} = a

= - cos^{2} (θ) + sin^{2} (θ)

Use the Double Angle identity:

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

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

= - cos (2 θ)

= - cos (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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