Is cos^4(θ)-sin^4(θ)=cos(2θ) ?

The answer to whether cos^4(θ)-sin^4(θ)=cos(2θ) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove cos^4(θ)-sin^4(θ)=cos(2θ)

prove

cos^{4} (θ) - sin^{4} (θ) = cos (2 θ)

True

Solution steps

cos^{4} (θ) - sin^{4} (θ) = cos (2 θ)

Manipulating left side

cos^{4} (θ) - sin^{4} (θ)

Factor

cos^{4} (θ) - sin^{4} (θ) : (cos^{2} (θ) + sin^{2} (θ)) (cos (θ) + sin (θ)) (cos (θ) - sin (θ))

cos^{4} (θ) - sin^{4} (θ)

Rewrite

cos^{4} (θ) - sin^{4} (θ)

(cos^{2} (θ))^{2} - (sin^{2} (θ))^{2}

cos^{4} (θ) - sin^{4} (θ)

Apply exponent rule:

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

sin^{4} (θ) = (sin^{2} (θ))^{2}

= cos^{4} (θ) - (sin^{2} (θ))^{2}

Apply exponent rule:

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

cos^{4} (θ) = (cos^{2} (θ))^{2}

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

Factor

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

cos^{2} (θ) - sin^{2} (θ)

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (cos (θ) + sin (θ)) (cos (θ) - sin (θ))

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= (cos (θ) + sin (θ)) (cos (θ) - sin (θ)) \cdot 1

Simplify

= (cos (θ) + sin (θ)) (cos (θ) - sin (θ))

Expand

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

(cos (θ) + sin (θ)) (cos (θ) - sin (θ))

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = cos (θ), b = sin (θ)

= cos^{2} (θ) - sin^{2} (θ)

= cos^{2} (θ) - sin^{2} (θ)

Use the Double Angle identity:

cos^{2} (θ) - sin^{2} (θ) = cos (2 θ)

= cos (2 θ)

= cos (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

cot^{2} (y) - cos^{2} (y) = cot^{2} (y) cos^{2} (y)

cot^{2} (x) = (csc (x) - 1) (csc (x) + 1)

sin (θ) - sin (θ) cos^{2} (θ) = sin^{3} (θ)

sin (x + \frac{π}{2}) = cos (x)

csc^{4} (x) - cot^{4} (x) = 2 csc^{2} (x) - 1

Is cos^4(θ)-sin^4(θ)=cos(2θ) ?
The answer to whether cos^4(θ)-sin^4(θ)=cos(2θ) is True