Is cos(3t)=4cos^3(t)-3cos(t) ?

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

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prove cos(3t)=4cos^3(t)-3cos(t)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

cos (3 t)

Rewrite using trig identities

cos (3 t)

Rewrite as

= cos (2 t + t)

Use the Angle Sum identity:

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

= cos (2 t) cos (t) - sin (2 t) sin (t)

Use the Double Angle identity:

sin (2 t) = 2 sin (t) cos (t)

= cos (2 t) cos (t) - 2 sin (t) cos (t) sin (t)

Simplify

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

cos (2 t) cos (t) - 2 sin (t) cos (t) sin (t)

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

2 sin (t) cos (t) sin (t)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

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

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

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

Expand

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

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Simplify

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

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

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

2 cos^{2} (t) cos (t)

Apply exponent rule:

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

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

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

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (t)

1 \cdot cos (t) = cos (t)

1 cos (t)

Multiply:

1 \cdot cos (t) = cos (t)

= cos (t)

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

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Apply minus-plus rules

- (- a) = a

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

Simplify

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

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

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

2 \cdot 1 cos (t)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos (t)

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

2 cos^{2} (t) cos (t)

Apply exponent rule:

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

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

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

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (t)

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add similar elements:

- cos (t) - 2 cos (t) = - 3 cos (t)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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