Is cos(2α)=-1+2/(1+tan^2(α)) ?

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

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prove cos(2α)=-1+2/(1+tan^2(α))

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

- 1 + \frac{2}{1 + tan ^{2} ( α )}

Express with sin, cos

- 1 + \frac{2}{1 + tan ^{2} ( α )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

\frac{2}{1 + ( \frac{s i n ( α )}{c o s ( α )} ) ^{2}} = \frac{2 cos ^{2} ( α )}{cos ^{2} ( α ) + sin ^{2} ( α )}

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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} ( α ) + sin ^{2} ( α )}{cos ^{2} ( α )}

Multiply:

1 \cdot cos^{2} (α) = cos^{2} (α)

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

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

Apply the fraction rule:

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

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

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

Convert element to fraction:

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

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

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} ( α ) + sin ^{2} ( α ) ) + 2 cos ^{2} ( α )}{cos ^{2} ( α ) + sin ^{2} ( α )}

Multiply:

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

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

Expand

- (cos^{2} (α) + sin^{2} (α)) + 2 cos^{2} (α) : cos^{2} (α) - sin^{2} (α)

- (cos^{2} (α) + sin^{2} (α)) + 2 cos^{2} (α)

- (cos^{2} (α) + sin^{2} (α)) : - cos^{2} (α) - sin^{2} (α)

- (cos^{2} (α) + sin^{2} (α))

Distribute parentheses

= - (cos^{2} (α)) - (sin^{2} (α))

Apply minus-plus rules

+ (- a) = - a

= - cos^{2} (α) - sin^{2} (α)

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

Add similar elements:

- cos^{2} (α) + 2 cos^{2} (α) = cos^{2} (α)

= cos^{2} (α) - sin^{2} (α)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

Apply rule

\frac{a}{1} = a

= cos^{2} (α) - sin^{2} (α)

Use the Double Angle identity:

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

= cos (2 α)

= cos (2 α)

We showed that the two sides could take the same form

\Rightarrow True

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