Is cos(4θ)=1-8sin^2(θ)cos^2(θ) ?

The answer to whether cos(4θ)=1-8sin^2(θ)cos^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(4θ)=1-8sin^2(θ)cos^2(θ)

prove

cos (4 θ) = 1 - 8 sin^{2} (θ) cos^{2} (θ)

True

Solution steps

cos (4 θ) = 1 - 8 sin^{2} (θ) cos^{2} (θ)

Manipulating left side

cos (4 θ)

Rewrite using trig identities

cos (4 θ)

= cos (2 \cdot 2 θ)

Use the Double Angle identity:

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

= 1 - 2 sin^{2} (2 θ)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= 1 - 2 (2 sin (θ) cos (θ))^{2}

Simplify

1 - 2 (2 sin (θ) cos (θ))^{2} : 1 - 8 sin^{2} (θ) cos^{2} (θ)

1 - 2 (2 sin (θ) cos (θ))^{2}

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

2 (2 sin (θ) cos (θ))^{2}

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

(2 sin (θ) cos (θ))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} sin^{2} (θ) cos^{2} (θ)

= 2^{2} \cdot 2 sin^{2} (θ) cos^{2} (θ)

Apply exponent rule:

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

2 \cdot 2^{2} = 2^{1 + 2}

= 2^{1 + 2} sin^{2} (θ) cos^{2} (θ)

Add the numbers:

1 + 2 = 3

= 2^{3} sin^{2} (θ) cos^{2} (θ)

= 1 - 2^{3} sin^{2} (θ) cos^{2} (θ)

2^{3} = 8

= 1 - 8 sin^{2} (θ) cos^{2} (θ)

= 1 - 8 sin^{2} (θ) cos^{2} (θ)

= 1 - 8 sin^{2} (θ) cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

cos^{6} (10 5^{\circ}) = (\frac{1 + cos ( 21 0 ^{\circ} )}{2})^{3}

cos^{2} (θ) = sin^{2} (θ) - 1

\frac{tan ( 2 θ ) + cot ( 2 θ )}{sec ( 2 θ )} = csc (2 θ)

\frac{sec ( θ )}{sec ( θ ) - 1} = \frac{1}{1 - cos ( θ )}

Is cos(4θ)=1-8sin^2(θ)cos^2(θ) ?
The answer to whether cos(4θ)=1-8sin^2(θ)cos^2(θ) is True