Is cos(3θ)=cos^3(θ)-3cos(θ)sin^2(θ) ?

The answer to whether cos(3θ)=cos^3(θ)-3cos(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

prove cos(3θ)=cos^3(θ)-3cos(θ)sin^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

cos (3 θ)

Rewrite using trig identities

cos (3 θ)

Rewrite as

= cos (2 θ + θ)

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (2 θ) cos (θ) - sin (2 θ) sin (θ)

Use the Double Angle identity:

sin (2 θ) = 2 sin (θ) cos (θ)

= cos (2 θ) cos (θ) - 2 sin (θ) cos (θ) sin (θ)

Simplify

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

cos (2 θ) cos (θ) - 2 sin (θ) cos (θ) sin (θ)

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

2 sin (θ) cos (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= 2 cos (θ) sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 2 cos (θ) sin^{2} (θ)

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

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

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

Use the Double Angle identity:

cos (2 θ) = cos^{2} (θ) - sin^{2} (θ)

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

Expand

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

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

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

Expand

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

cos (θ) (cos^{2} (θ) - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = cos (θ), b = cos^{2} (θ), c = sin^{2} (θ)

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

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

cos^{2} (θ) cos (θ) = cos^{3} (θ)

cos^{2} (θ) cos (θ)

Apply exponent rule:

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

cos^{2} (θ) cos (θ) = cos^{2 + 1} (θ)

= cos^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= cos^{3} (θ)

= cos^{3} (θ) - sin^{2} (θ) cos (θ)

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

Add similar elements:

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

= cos^{3} (θ) - 3 sin^{2} (θ) cos (θ)

= cos^{3} (θ) - 3 sin^{2} (θ) cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{tan ^{2} ( x )}{sec ( x )} = sin (x) tan (x)

p ro v e sec (u) - tan (u) = \frac{cos ( u )}{1 + sin ( u )}

p ro v e cot (x) + \frac{sin ( x )}{1 + cos ( x )} = csc (x)

p ro v e \frac{csc ^{2} ( x )}{2 cot ( x )} = csc (2 x)

p ro v e cot (x) (sec^{2} (x) - 1) = tan (x)

Is cos(3θ)=cos^3(θ)-3cos(θ)sin^2(θ) ?
The answer to whether cos(3θ)=cos^3(θ)-3cos(θ)sin^2(θ) is True