Is 2sin(a+b)sin(a-b)=cos(2b)-cos(2a) ?

The answer to whether 2sin(a+b)sin(a-b)=cos(2b)-cos(2a) is True

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prove 2sin(a+b)sin(a-b)=cos(2b)-cos(2a)

prove

2 sin (a + b) sin (a - b) = cos (2 b) - cos (2 a)

True

Solution steps

2 sin (a + b) sin (a - b) = cos (2 b) - cos (2 a)

Manipulating left side

2 sin (a + b) sin (a - b)

Rewrite using trig identities

2 sin (a + b) sin (a - b)

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 (a + b - (a - b)) - cos (a + b + a - b))

Simplify

2 \cdot \frac{1}{2} (cos (a + b - (a - b)) - cos (a + b + a - b)) : cos (2 b) - cos (2 a)

2 \cdot \frac{1}{2} (cos (a + b - (a - b)) - cos (a + b + a - b))

Multiply fractions:

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

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

Cancel the common factor:

2

= 1 \cdot (cos (a + b - (a - b)) - cos (a + b + a - b))

Expand

a + b - (a - b) : 2 b

a + b - (a - b)

- (a - b) : - a + b

- (a - b)

Distribute parentheses

= - (a) - (- b)

Apply minus-plus rules

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

= - a + b

= a + b - a + b

Simplify

a + b - a + b : 2 b

a + b - a + b

Group like terms

= a - a + b + b

Add similar elements:

a - a = 0

= b + b

Add similar elements:

b + b = 2 b

= 2 b

= 2 b

= 1 \cdot (cos (2 b) - cos (a + a + b - b))

Simplify

a + b + a - b : 2 a

a + b + a - b

Group like terms

= a + a + b - b

Add similar elements:

a + a = 2 a

= 2 a + b - b

Add similar elements:

b - b = 0

= 2 a

= 1 \cdot (cos (2 b) - cos (2 a))

Refine

= cos (2 b) - cos (2 a)

= cos (2 b) - cos (2 a)

= cos (2 b) - cos (2 a)

We showed that the two sides could take the same form

\Rightarrow True

\frac{1}{sin ( x )} = 2

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

(4 sin (θ) + 4 cos (θ)) = 16 + 16 sin (2 θ)

5 sec^{2} (x) = 5 sec^{2} (- x)

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

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