Is sin^3(θ)= 3/4 sin(θ)-1/4 sin(3θ) ?

The answer to whether sin^3(θ)= 3/4 sin(θ)-1/4 sin(3θ) is True

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prove sin^3(θ)= 3/4 sin(θ)-1/4 sin(3θ)

Solution

prove

sin^{3} (θ) = \frac{3}{4} sin (θ) - \frac{1}{4} sin (3 θ)

Solution

True

Solution steps

sin^{3} (θ) = \frac{3}{4} sin (θ) - \frac{1}{4} sin (3 θ)

Manipulating right side

\frac{3}{4} sin (θ) - \frac{1}{4} sin (3 θ)

Rewrite using trig identities

\frac{3}{4} sin (θ) - \frac{1}{4} sin (3 θ)

Use the following identity:

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

sin (3 x)

Rewrite using trig identities

sin (3 x)

Rewrite as

= sin (2 x + x)

Use the Angle Sum identity:

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

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

Use the Double Angle identity:

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

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

Simplify

cos (2 x) sin (x) + cos (x) \cdot 2 sin (x) cos (x) : sin (x) cos (2 x) + 2 cos^{2} (x) sin (x)

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

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

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

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

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

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

Use the Double Angle identity:

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

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

Use the Pythagorean identity:

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

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

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

Expand

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

(1 - 2 sin^{2} (x)) sin (x) + 2 (1 - sin^{2} (x)) sin (x)

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Simplify

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

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

1 \cdot sin (x) = sin (x)

1 sin (x)

Multiply:

1 \cdot sin (x) = sin (x)

= sin (x)

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

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

Apply exponent rule:

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

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

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

Add the numbers:

2 + 1 = 3

= 2 sin^{3} (x)

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

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

= sin (x) - 2 sin^{3} (x) + 2 (1 - sin^{2} (x)) sin (x)

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Simplify

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

2 \cdot 1 sin (x) - 2 sin^{2} (x) sin (x)

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

2 \cdot 1 sin (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 sin (x)

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

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

Apply exponent rule:

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

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

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

Add the numbers:

2 + 1 = 3

= 2 sin^{3} (x)

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

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add similar elements:

sin (x) + 2 sin (x) = 3 sin (x)

= - 4 sin^{3} (x) + 3 sin (x)

= - 4 sin^{3} (x) + 3 sin (x)

= - 4 sin^{3} (x) + 3 sin (x)

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

= \frac{3}{4} sin (θ) - \frac{1}{4} (3 sin (θ) - 4 sin^{3} (θ))

Simplify

\frac{3}{4} sin (θ) - \frac{1}{4} (3 sin (θ) - 4 sin^{3} (θ)) : sin^{3} (θ)

\frac{3}{4} sin (θ) - \frac{1}{4} (3 sin (θ) - 4 sin^{3} (θ))

Expand

- \frac{1}{4} (3 sin (θ) - 4 sin^{3} (θ)) : - \frac{3}{4} sin (θ) + sin^{3} (θ)

- \frac{1}{4} (3 sin (θ) - 4 sin^{3} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - \frac{1}{4}, b = 3 sin (θ), c = 4 sin^{3} (θ)

= - \frac{1}{4} \cdot 3 sin (θ) - (- \frac{1}{4}) \cdot 4 sin^{3} (θ)

Apply minus-plus rules

- (- a) = a

= - 3 \cdot \frac{1}{4} sin (θ) + 4 \cdot \frac{1}{4} sin^{3} (θ)

Simplify

- 3 \cdot \frac{1}{4} sin (θ) + 4 \cdot \frac{1}{4} sin^{3} (θ) : - \frac{3}{4} sin (θ) + sin^{3} (θ)

- 3 \cdot \frac{1}{4} sin (θ) + 4 \cdot \frac{1}{4} sin^{3} (θ)

3 \cdot \frac{1}{4} sin (θ) = \frac{3}{4} sin (θ)

3 \cdot \frac{1}{4} sin (θ)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 3}{4} sin (θ)

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{4} sin (θ)

4 \cdot \frac{1}{4} sin^{3} (θ) = sin^{3} (θ)

4 \cdot \frac{1}{4} sin^{3} (θ)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 4}{4} sin^{3} (θ)

Cancel the common factor:

4

= sin^{3} (θ) \cdot 1

Multiply:

sin^{3} (θ) \cdot 1 = sin^{3} (θ)

= sin^{3} (θ)

= - \frac{3}{4} sin (θ) + sin^{3} (θ)

= - \frac{3}{4} sin (θ) + sin^{3} (θ)

= \frac{3}{4} sin (θ) - \frac{3}{4} sin (θ) + sin^{3} (θ)

Add similar elements:

\frac{3}{4} sin (θ) - \frac{3}{4} sin (θ) = 0

\frac{3}{4} sin (θ) - \frac{3}{4} sin (θ)

Factor out common term

sin (θ)

= sin (θ) (\frac{3}{4} - \frac{3}{4})

\frac{3}{4} - \frac{3}{4} = 0

\frac{3}{4} - \frac{3}{4}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{3 - 3}{4}

Refine

= 0

= 0

= sin^{3} (θ)

= sin^{3} (θ)

= sin^{3} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is sin^3(θ)= 3/4 sin(θ)-1/4 sin(3θ) ?
The answer to whether sin^3(θ)= 3/4 sin(θ)-1/4 sin(3θ) is True