Is cos(3α)=4cos^3(α)-3cos(α) ?

The answer to whether cos(3α)=4cos^3(α)-3cos(α) is True

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Popular Trigonometry >

prove cos(3α)=4cos^3(α)-3cos(α)

Solution

prove

cos (3 α) = 4 cos^{3} (α) - 3 cos (α)

Solution

True

Solution steps

cos (3 α) = 4 cos^{3} (α) - 3 cos (α)

Manipulating left side

cos (3 α)

Rewrite using trig identities

cos (3 α)

Rewrite as

= cos (2 α + α)

Use the Angle Sum identity:

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

= cos (2 α) cos (α) - sin (2 α) sin (α)

Use the Double Angle identity:

sin (2 α) = 2 sin (α) cos (α)

= cos (2 α) cos (α) - 2 sin (α) cos (α) sin (α)

Simplify

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

cos (2 α) cos (α) - 2 sin (α) cos (α) sin (α)

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

2 sin (α) cos (α) sin (α)

Apply exponent rule:

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

sin (α) sin (α) = sin^{1 + 1} (α)

= 2 cos (α) sin^{1 + 1} (α)

Add the numbers:

1 + 1 = 2

= 2 cos (α) sin^{2} (α)

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

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

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

Use the Double Angle identity:

cos (2 α) = 2 cos^{2} (α) - 1

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

Use the Pythagorean identity:

cos^{2} (α) + sin^{2} (α) = 1

sin^{2} (α) = 1 - cos^{2} (α)

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

Expand

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

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

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

Expand

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

cos (α) (2 cos^{2} (α) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = cos (α), b = 2 cos^{2} (α), c = 1

= cos (α) 2 cos^{2} (α) - cos (α) 1

= 2 cos^{2} (α) cos (α) - 1 cos (α)

Simplify

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

2 cos^{2} (α) cos (α) - 1 cos (α)

2 cos^{2} (α) cos (α) = 2 cos^{3} (α)

2 cos^{2} (α) cos (α)

Apply exponent rule:

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

cos^{2} (α) cos (α) = cos^{2 + 1} (α)

= 2 cos^{2 + 1} (α)

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (α)

1 \cdot cos (α) = cos (α)

1 cos (α)

Multiply:

1 \cdot cos (α) = cos (α)

= cos (α)

= 2 cos^{3} (α) - cos (α)

= 2 cos^{3} (α) - cos (α)

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

Expand

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

- 2 cos (α) (1 - cos^{2} (α))

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Simplify

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

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

2 \cdot 1 \cdot cos (α) = 2 cos (α)

2 \cdot 1 cos (α)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos (α)

2 cos^{2} (α) cos (α) = 2 cos^{3} (α)

2 cos^{2} (α) cos (α)

Apply exponent rule:

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

cos^{2} (α) cos (α) = cos^{2 + 1} (α)

= 2 cos^{2 + 1} (α)

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (α)

= - 2 cos (α) + 2 cos^{3} (α)

= - 2 cos (α) + 2 cos^{3} (α)

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

= 4 cos^{3} (α) - cos (α) - 2 cos (α)

Add similar elements:

- cos (α) - 2 cos (α) = - 3 cos (α)

= 4 cos^{3} (α) - 3 cos (α)

= 4 cos^{3} (α) - 3 cos (α)

= 4 cos^{3} (α) - 3 cos (α)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is cos(3α)=4cos^3(α)-3cos(α) ?
The answer to whether cos(3α)=4cos^3(α)-3cos(α) is True