Is sin(3x)=(sin(x))(4cos^2(x)-1) ?

The answer to whether sin(3x)=(sin(x))(4cos^2(x)-1) is True

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Popular Trigonometry >

prove sin(3x)=(sin(x))(4cos^2(x)-1)

Solution

prove

sin (3 x) = (sin (x)) (4 cos^{2} (x) - 1)

Solution

True

Solution steps

sin (3 x) = sin (x) (4 cos^{2} (x) - 1)

Manipulating left side

sin (3 x)

Rewrite using trig identities

sin (3 x)

Use the following identity:

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

Use the Double Angle identity:

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

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

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

Expand

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

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

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

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

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

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

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)

= sin^{2 + 1} (x)

Add the numbers:

2 + 1 = 3

= sin^{3} (x)

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

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

Simplify

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

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

Group like terms

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

Add similar elements:

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

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

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

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

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

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

Factor

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

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

Apply exponent rule:

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

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

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

Factor out common term

sin (x)

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

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

Manipulating right side

sin (x) (4 cos^{2} (x) - 1)

Rewrite using trig identities

sin (x) (4 cos^{2} (x) - 1)

Use the Pythagorean identity:

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

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

Add similar elements:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is sin(3x)=(sin(x))(4cos^2(x)-1) ?
The answer to whether sin(3x)=(sin(x))(4cos^2(x)-1) is True