Is cos(α-β)-cos(α+β)=2sin(α)sin(β) ?

The answer to whether cos(α-β)-cos(α+β)=2sin(α)sin(β) is True

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prove cos(α-β)-cos(α+β)=2sin(α)sin(β)

prove

cos (α - β) - cos (α + β) = 2 sin (α) sin (β)

True

Solution steps

cos (α - β) - cos (α + β) = 2 sin (α) sin (β)

Manipulating right side

2 sin (α) sin (β)

Rewrite using trig identities

2 sin (α) sin (β)

Use the Product to Sum identity:

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

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

Simplify

2 \cdot \frac{1}{2} (cos (α - β) - cos (α + β)) : cos (α - β) - cos (α + β)

2 \cdot \frac{1}{2} (cos (α - β) - cos (α + β))

Multiply fractions:

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

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

Cancel the common factor:

2

= (cos (α - β) - cos (α + β)) \cdot 1

Refine

= cos (α - β) - cos (α + β)

= cos (α - β) - cos (α + β)

= cos (α - β) - cos (α + β)

We showed that the two sides could take the same form

\Rightarrow True

tan (x) - 10 = \frac{sin ( x ) + 10}{cos ( x )}

csc^{2} (x) = \frac{sec ^{2} ( x )}{tan ^{2} ( x )}

sin (x) cos (x) = \frac{1}{tan ( x ) + cot ( x )}

1 = sec^{2} (θ) - tan^{2} (θ)

\frac{1}{sin ( θ ) cot ( θ )} = \frac{1}{cos ( θ )}

Is cos(α-β)-cos(α+β)=2sin(α)sin(β) ?
The answer to whether cos(α-β)-cos(α+β)=2sin(α)sin(β) is True