Is (2cos(θ))^3=2cos(3θ)+6cos(θ) ?

The answer to whether (2cos(θ))^3=2cos(3θ)+6cos(θ) is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove (2cos(θ))^3=2cos(3θ)+6cos(θ)

Solution

prove

(2 cos (θ))^{3} = 2 cos (3 θ) + 6 cos (θ)

Solution

True

Solution steps

(2 cos (θ))^{3} = 2 cos (3 θ) + 6 cos (θ)

Manipulating left side

(2 cos (θ))^{3}

Simplify

(2 cos (θ))^{3} : 8 cos^{3} (θ)

(2 cos (θ))^{3}

Apply exponent rule:

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

= 2^{3} cos^{3} (θ)

2^{3} = 8

= 8 cos^{3} (θ)

= 8 cos^{3} (θ)

Manipulating right side

2 cos (3 θ) + 6 cos (θ)

Rewrite using trig identities

2 cos (3 θ) + 6 cos (θ)

Use the following identity:

cos (3 x) = 4 cos^{3} (x) - 3 cos (x)

cos (3 x)

Rewrite using trig identities

cos (3 x)

Rewrite as

= cos (2 x + x)

Use the Angle Sum identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

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

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

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

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

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

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

Expand

(2 cos^{2} (x) - 1) cos (x) - 2 (1 - cos^{2} (x)) cos (x) : 4 cos^{3} (x) - 3 cos (x)

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

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

Expand

cos (x) (2 cos^{2} (x) - 1) : 2 cos^{3} (x) - cos (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = cos (x), b = 2 cos^{2} (x), c = 1

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

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

Simplify

2 cos^{2} (x) cos (x) - 1 \cdot cos (x) : 2 cos^{3} (x) - cos (x)

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

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

2 cos^{2} (x) cos (x)

Apply exponent rule:

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

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

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

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (x)

1 \cdot cos (x) = cos (x)

1 cos (x)

Multiply:

1 \cdot cos (x) = cos (x)

= cos (x)

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

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

= 2 cos^{3} (x) - cos (x) - 2 (1 - cos^{2} (x)) cos (x)

Expand

- 2 cos (x) (1 - cos^{2} (x)) : - 2 cos (x) + 2 cos^{3} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = - 2 cos (x), b = 1, c = cos^{2} (x)

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

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 cos (x) + 2 cos^{2} (x) cos (x)

Simplify

- 2 \cdot 1 \cdot cos (x) + 2 cos^{2} (x) cos (x) : - 2 cos (x) + 2 cos^{3} (x)

- 2 \cdot 1 cos (x) + 2 cos^{2} (x) cos (x)

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 cos (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos (x)

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

2 cos^{2} (x) cos (x)

Apply exponent rule:

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

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

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

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= - 2 cos (x) + 2 cos^{3} (x)

= 2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

Simplify

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x) : 4 cos^{3} (x) - 3 cos (x)

2 cos^{3} (x) - cos (x) - 2 cos (x) + 2 cos^{3} (x)

Group like terms

= 2 cos^{3} (x) + 2 cos^{3} (x) - cos (x) - 2 cos (x)

Add similar elements:

2 cos^{3} (x) + 2 cos^{3} (x) = 4 cos^{3} (x)

= 4 cos^{3} (x) - cos (x) - 2 cos (x)

Add similar elements:

- cos (x) - 2 cos (x) = - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 4 cos^{3} (x) - 3 cos (x)

= 2 (4 cos^{3} (θ) - 3 cos (θ)) + 6 cos (θ)

Simplify

2 (4 cos^{3} (θ) - 3 cos (θ)) + 6 cos (θ) : 8 cos^{3} (θ)

2 (4 cos^{3} (θ) - 3 cos (θ)) + 6 cos (θ)

Expand

2 (4 cos^{3} (θ) - 3 cos (θ)) : 8 cos^{3} (θ) - 6 cos (θ)

2 (4 cos^{3} (θ) - 3 cos (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 4 cos^{3} (θ), c = 3 cos (θ)

= 2 \cdot 4 cos^{3} (θ) - 2 \cdot 3 cos (θ)

Simplify

2 \cdot 4 cos^{3} (θ) - 2 \cdot 3 cos (θ) : 8 cos^{3} (θ) - 6 cos (θ)

2 \cdot 4 cos^{3} (θ) - 2 \cdot 3 cos (θ)

Multiply the numbers:

2 \cdot 4 = 8

= 8 cos^{3} (θ) - 2 \cdot 3 cos (θ)

Multiply the numbers:

2 \cdot 3 = 6

= 8 cos^{3} (θ) - 6 cos (θ)

= 8 cos^{3} (θ) - 6 cos (θ)

= 8 cos^{3} (θ) - 6 cos (θ) + 6 cos (θ)

Add similar elements:

- 6 cos (θ) + 6 cos (θ) = 0

= 8 cos^{3} (θ)

= 8 cos^{3} (θ)

= 8 cos^{3} (θ)

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove sin(x)tan(x)=cos(x)prove

sin (x) tan (x) = cos (x)

prove 2sin^2(x/2)+4cos^2(x/2)-3=cos(x)prove

2 sin^{2} (\frac{x}{2}) + 4 cos^{2} (\frac{x}{2}) - 3 = cos (x)

prove tan(2u)=((2tan(u)))/(1-tan^2(u))prove

tan (2 u) = \frac{( 2 tan ( u ) )}{1 - tan ^{2} ( u )}

prove sin(4x)=2sin(2x)prove

sin (4 x) = 2 sin (2 x)

prove 1/(-sin(x))=sec(x)tan(x)prove

\frac{1}{- sin ( x )} = sec (x) tan (x)

Frequently Asked Questions (FAQ)

Is (2cos(θ))^3=2cos(3θ)+6cos(θ) ?
The answer to whether (2cos(θ))^3=2cos(3θ)+6cos(θ) is True