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

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

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

prove

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

True

Solution steps

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

Manipulating right side

2 cos (x) sin (b)

Rewrite using trig identities

2 cos (x) sin (b)

Use the Product to Sum identity:

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

= 2 \cdot \frac{1}{2} (sin (x + b) - sin (x - b))

Simplify

2 \cdot \frac{1}{2} (sin (x + b) - sin (x - b)) : sin (x + b) - sin (x - b)

2 \cdot \frac{1}{2} (sin (x + b) - sin (x - b))

Multiply fractions:

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

= \frac{1 \cdot 2}{2} (sin (x + b) - sin (x - b))

Cancel the common factor:

2

= (sin (x + b) - sin (x - b)) \cdot 1

Refine

= sin (x + b) - sin (x - b)

= sin (x + b) - sin (x - b)

= sin (x + b) - sin (x - b)

We showed that the two sides could take the same form

\Rightarrow True

sin (4 x) = 2 sin (x) (cos (x) + cos (3 x))

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

sec^{2} (θ) - cos^{2} (θ) = tan (θ) sin (θ)

\frac{1}{sec ( θ ) - tan ( θ )} = sec (θ) tan (θ)

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

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