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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

prove

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

True

Solution steps

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

Manipulating left side

cos^{2} (x) (1 - tan^{2} (x))

Express with sin, cos

(1 - tan^{2} (x)) cos^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

= cos (2 x)

= cos (2 x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

cos^{3} (2 x) = (1 - sin^{2} (2 x)) cos (2 x)

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

tan (18 0^{\circ} - y) = - tan (y)

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