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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

a = 0

For

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

plug in

b = 1

sin (0 + 1) - sin (0 - 1) = 1.68294 \dots

sin (0 + 1) - sin (0 - 1)

Refine to a decimal form

= 1.68294 \dots

2 sin (0) sin (1) = 0

2 sin (0) sin (1)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 2 \cdot 0 \cdot sin (1)

= 2 \cdot 0 \cdot sin (1)

Simplify

= 0

The two sides are not equal

\Rightarrow False

sec (x) + 1 = \frac{tan ^{2} ( x )}{sec ( x ) - 1}

\frac{1}{cot ^{2} ( x )} + sec (x) cos (x) = csc^{2} (x)

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

1 csc (θ) - (cos (θ)) (cot (θ)) = sin (θ)

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

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