Is cos(θ)-sin(θ)sin(2θ)=cos(θ)cos(2θ) ?

The answer to whether cos(θ)-sin(θ)sin(2θ)=cos(θ)cos(2θ) is True

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prove cos(θ)-sin(θ)sin(2θ)=cos(θ)cos(2θ)

Solution

prove

cos (θ) - sin (θ) sin (2 θ) = cos (θ) cos (2 θ)

Solution

True

Solution steps

cos (θ) - sin (θ) sin (2 θ) = cos (θ) cos (2 θ)

Manipulating left side

cos (θ) - sin (θ) sin (2 θ)

Rewrite using trig identities

cos (θ) - sin (θ) sin (2 θ)

Use the Product to Sum identity:

sin (s) sin (t) = \frac{1}{2} (cos (s - t) - cos (s + t))

= cos (θ) - \frac{1}{2} (cos (θ - 2 θ) - cos (θ + 2 θ))

Simplify

cos (θ) - \frac{1}{2} (cos (θ - 2 θ) - cos (θ + 2 θ)) : \frac{cos ( θ ) + cos ( 3 θ )}{2}

cos (θ) - \frac{1}{2} (cos (θ - 2 θ) - cos (θ + 2 θ))

\frac{1}{2} (cos (θ - 2 θ) - cos (θ + 2 θ)) = \frac{cos ( θ ) - cos ( 3 θ )}{2}

\frac{1}{2} (cos (θ - 2 θ) - cos (θ + 2 θ))

Multiply fractions:

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

= \frac{1 \cdot ( cos ( θ - 2 θ ) - cos ( θ + 2 θ ) )}{2}

1 \cdot (cos (θ - 2 θ) - cos (θ + 2 θ)) = cos (- θ) - cos (3 θ)

1 \cdot (cos (θ - 2 θ) - cos (θ + 2 θ))

Add similar elements:

θ - 2 θ = - θ

= 1 \cdot (cos (- θ) - cos (θ + 2 θ))

Add similar elements:

θ + 2 θ = 3 θ

= 1 \cdot (cos (- θ) - cos (3 θ))

Multiply:

1 \cdot (cos (- θ) - cos (3 θ)) = (cos (- θ) - cos (3 θ))

= (cos (- θ) - cos (3 θ))

Remove parentheses:

(a) = a

= cos (- θ) - cos (3 θ)

= \frac{cos ( - θ ) - cos ( 3 θ )}{2}

Simplify

cos (- θ) - cos (3 θ) : cos (θ) - cos (3 θ)

cos (- θ) - cos (3 θ)

Use the negative angle identity:

cos (- x) = cos (x)

= cos (θ) - cos (3 θ)

= \frac{cos ( θ ) - cos ( 3 θ )}{2}

= cos (θ) - \frac{cos ( θ ) - cos ( 3 θ )}{2}

Convert element to fraction:

cos (θ) = \frac{cos ( θ ) 2}{2}

= - \frac{cos ( θ ) - cos ( 3 θ )}{2} + \frac{cos ( θ ) \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{- ( cos ( θ ) - cos ( 3 θ ) ) + cos ( θ ) \cdot 2}{2}

Expand

- (cos (θ) - cos (3 θ)) + cos (θ) \cdot 2 : cos (θ) + cos (3 θ)

- (cos (θ) - cos (3 θ)) + cos (θ) \cdot 2

= - (cos (θ) - cos (3 θ)) + 2 cos (θ)

- (cos (θ) - cos (3 θ)) : - cos (θ) + cos (3 θ)

- (cos (θ) - cos (3 θ))

Distribute parentheses

= - (cos (θ)) - (- cos (3 θ))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - cos (θ) + cos (3 θ)

= - cos (θ) + cos (3 θ) + cos (θ) \cdot 2

Simplify

- cos (θ) + cos (3 θ) + cos (θ) \cdot 2 : cos (θ) + cos (3 θ)

- cos (θ) + cos (3 θ) + cos (θ) \cdot 2

Group like terms

= - cos (θ) + cos (3 θ) + 2 cos (θ)

Add similar elements:

- cos (θ) + 2 cos (θ) = cos (θ)

= cos (θ) + cos (3 θ)

= cos (θ) + cos (3 θ)

= \frac{cos ( θ ) + cos ( 3 θ )}{2}

= \frac{cos ( θ ) + cos ( 3 θ )}{2}

= \frac{cos ( θ ) + cos ( 3 θ )}{2}

Manipulating right side

cos (θ) cos (2 θ)

Rewrite using trig identities

cos (θ) cos (2 θ)

Use the Product to Sum identity:

cos (s) cos (t) = \frac{1}{2} (cos (s - t) + cos (s + t))

= \frac{1}{2} (cos (θ - 2 θ) + cos (θ + 2 θ))

Simplify

\frac{1}{2} (cos (θ - 2 θ) + cos (θ + 2 θ)) : \frac{cos ( θ ) + cos ( 3 θ )}{2}

\frac{1}{2} (cos (θ - 2 θ) + cos (θ + 2 θ))

Multiply fractions:

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

= \frac{1 \cdot ( cos ( θ - 2 θ ) + cos ( θ + 2 θ ) )}{2}

1 \cdot (cos (θ - 2 θ) + cos (θ + 2 θ)) = cos (- θ) + cos (3 θ)

1 \cdot (cos (θ - 2 θ) + cos (θ + 2 θ))

Add similar elements:

θ - 2 θ = - θ

= 1 \cdot (cos (- θ) + cos (θ + 2 θ))

Add similar elements:

θ + 2 θ = 3 θ

= 1 \cdot (cos (- θ) + cos (3 θ))

Multiply:

1 \cdot (cos (- θ) + cos (3 θ)) = (cos (- θ) + cos (3 θ))

= (cos (- θ) + cos (3 θ))

Remove parentheses:

(a) = a

= cos (- θ) + cos (3 θ)

= \frac{cos ( - θ ) + cos ( 3 θ )}{2}

Simplify

cos (- θ) + cos (3 θ) : cos (θ) + cos (3 θ)

cos (- θ) + cos (3 θ)

Use the negative angle identity:

cos (- x) = cos (x)

= cos (θ) + cos (3 θ)

= \frac{cos ( θ ) + cos ( 3 θ )}{2}

= \frac{cos ( θ ) + cos ( 3 θ )}{2}

= \frac{cos ( θ ) + cos ( 3 θ )}{2}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is cos(θ)-sin(θ)sin(2θ)=cos(θ)cos(2θ) ?
The answer to whether cos(θ)-sin(θ)sin(2θ)=cos(θ)cos(2θ) is True