Is sin(3θ)=3sin(θ)-4sin^3(θ) ?

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

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prove sin(3θ)=3sin(θ)-4sin^3(θ)

Solution

prove

sin (3 θ) = 3 sin (θ) - 4 sin^{3} (θ)

Solution

True

Solution steps

sin (3 θ) = 3 sin (θ) - 4 sin^{3} (θ)

Manipulating left side

sin (3 θ)

Rewrite using trig identities

sin (3 θ)

Rewrite as

= sin (2 θ + θ)

Use the Angle Sum identity:

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

= sin (2 θ) cos (θ) + cos (2 θ) sin (θ)

Use the Double Angle identity:

sin (2 θ) = 2 sin (θ) cos (θ)

= cos (2 θ) sin (θ) + cos (θ) 2 sin (θ) cos (θ)

Simplify

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

cos (2 θ) sin (θ) + cos (θ) 2 sin (θ) cos (θ)

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

cos (θ) 2 sin (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= 2 sin (θ) cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 2 sin (θ) cos^{2} (θ)

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

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

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

Use the Double Angle identity:

cos (2 θ) = 1 - 2 sin^{2} (θ)

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

Use the Pythagorean identity:

cos^{2} (θ) + sin^{2} (θ) = 1

cos^{2} (θ) = 1 - sin^{2} (θ)

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

Expand

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

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

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

Expand

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

sin (θ) (1 - 2 sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = sin (θ), b = 1, c = 2 sin^{2} (θ)

= sin (θ) 1 - sin (θ) 2 sin^{2} (θ)

= 1 sin (θ) - 2 sin^{2} (θ) sin (θ)

Simplify

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

1 sin (θ) - 2 sin^{2} (θ) sin (θ)

1 \cdot sin (θ) = sin (θ)

1 sin (θ)

Multiply:

1 \cdot sin (θ) = sin (θ)

= sin (θ)

2 sin^{2} (θ) sin (θ) = 2 sin^{3} (θ)

2 sin^{2} (θ) sin (θ)

Apply exponent rule:

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

sin^{2} (θ) sin (θ) = sin^{2 + 1} (θ)

= 2 sin^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= 2 sin^{3} (θ)

= sin (θ) - 2 sin^{3} (θ)

= sin (θ) - 2 sin^{3} (θ)

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

Expand

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

2 sin (θ) (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 2 sin (θ), b = 1, c = sin^{2} (θ)

= 2 sin (θ) 1 - 2 sin (θ) sin^{2} (θ)

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

Simplify

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

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

2 \cdot 1 \cdot sin (θ) = 2 sin (θ)

2 \cdot 1 sin (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 sin (θ)

2 sin^{2} (θ) sin (θ) = 2 sin^{3} (θ)

2 sin^{2} (θ) sin (θ)

Apply exponent rule:

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

sin^{2} (θ) sin (θ) = sin^{2 + 1} (θ)

= 2 sin^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= 2 sin^{3} (θ)

= 2 sin (θ) - 2 sin^{3} (θ)

= 2 sin (θ) - 2 sin^{3} (θ)

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

Add similar elements:

sin (θ) + 2 sin (θ) = 3 sin (θ)

= - 4 sin^{3} (θ) + 3 sin (θ)

= - 4 sin^{3} (θ) + 3 sin (θ)

= - 4 sin^{3} (θ) + 3 sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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