Is 1-cot^2(x)sin^2(x)=sin^2(x) ?

The answer to whether 1-cot^2(x)sin^2(x)=sin^2(x) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove 1-cot^2(x)sin^2(x)=sin^2(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (x)

= cos^{2} (x)

= 1 - cos^{2} (x)

= 1 - cos^{2} (x)

= 1 - cos^{2} (x)

Rewrite using trig identities

1 - cos^{2} (x)

Use the Pythagorean identity:

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

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

= sin^{2} (x)

= sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

csc (ω) - sin (ω) = cot (ω) cos (ω)

\frac{cot ( 3 θ ) - tan ( 3 θ )}{sin ( 3 θ ) + cos ( 3 θ )} = csc (3 θ) - sec (3 θ)

sin (- \frac{π}{2} - θ) = - cos (θ)

- 2 cos (2 x) \cdot sin (2 x) = sin (- 4 x)

sec^{2} (x) + tan (x) - 1 = 0

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