Is cos(3β)=cos(β)(cos^2(β)-3sin^2(β)) ?

The answer to whether cos(3β)=cos(β)(cos^2(β)-3sin^2(β)) is True

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Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

cos (3 β)

Rewrite using trig identities

cos (3 β)

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)

= 4 cos^{3} (β) - 3 cos (β)

= 4 cos^{3} (β) - 3 cos (β)

Factor

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

- 3 cos (β) + 4 cos^{3} (β)

Apply exponent rule:

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

cos^{3} (β) = cos (β) cos^{2} (β)

= - 3 cos (β) + 4 cos (β) cos^{2} (β)

Factor out common term

cos (β)

= cos (β) (- 3 + 4 cos^{2} (β))

= (- 3 + 4 cos^{2} (β)) cos (β)

Rewrite using trig identities

(- 3 + 4 cos^{2} (β)) cos (β)

Use the Pythagorean identity:

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

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

= (- 3 + 4 (1 - sin^{2} (β))) cos (β)

Expand

- 3 + 4 (1 - sin^{2} (β)) : - 4 sin^{2} (β) + 1

- 3 + 4 (1 - sin^{2} (β))

Expand

4 (1 - sin^{2} (β)) : 4 - 4 sin^{2} (β)

4 (1 - sin^{2} (β))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (β)

= 4 \cdot 1 - 4 sin^{2} (β)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 sin^{2} (β)

= - 3 + 4 - 4 sin^{2} (β)

Add/Subtract the numbers:

- 3 + 4 = 1

= - 4 sin^{2} (β) + 1

= cos (β) (- 4 sin^{2} (β) + 1)

Use the Pythagorean identity:

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

= cos (β) (- 4 sin^{2} (β) + cos^{2} (β) + sin^{2} (β))

Add similar elements:

- 4 sin^{2} (β) + sin^{2} (β) = - 3 sin^{2} (β)

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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