Is cos(θ)tan(θ)sin(θ)=sin^2(θ) ?

The answer to whether cos(θ)tan(θ)sin(θ)=sin^2(θ) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove cos(θ)tan(θ)sin(θ)=sin^2(θ)

prove

cos (θ) tan (θ) sin (θ) = sin^{2} (θ)

True

Solution steps

cos (θ) tan (θ) sin (θ) = sin^{2} (θ)

Manipulating left side

cos (θ) tan (θ) sin (θ)

Express with sin, cos

cos (θ) sin (θ) tan (θ)

Use the basic trigonometric identity:

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

= cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )}

Simplify

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

cos (θ) sin (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) cos ( θ ) sin ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= sin^{2} (θ)

= sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

cos (\frac{3 x}{5}) = - sin (\frac{3 x}{5} - \frac{π}{2})

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

\frac{( cos ( a - b ) )}{sin ( a ) sin ( b )} = cot (a) cot (b) + 1

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

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

Is cos(θ)tan(θ)sin(θ)=sin^2(θ) ?
The answer to whether cos(θ)tan(θ)sin(θ)=sin^2(θ) is True