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

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

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

prove

sin (a) sin (b) = sin (a + b)

False

Solution steps

sin (a) sin (b) = sin (a + b)

Show that the two sides are not equal

For

sin (a) sin (b) = sin (a + b)

plug in

a = 0

For

sin (a) sin (b) = sin (a + b)

plug in

b = 1

sin (0) sin (1) = 0

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}

= 0 \cdot sin (1)

= 0 \cdot sin (1)

Simplify

= 0

sin (0 + 1) = 0.84147 \dots

sin (0 + 1)

Refine to a decimal form

= 0.84147 \dots

The two sides are not equal

\Rightarrow False

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

2 sin (x) = \frac{( 4 cos ( x ) - 1 )}{tan ( x )}

\frac{( 1 - cos ( 4 x ) + sin ( 4 x ) )}{( 1 + cos ( 4 x ) + sin ( 4 x ) )} = 3 sin (2 x)

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

3 tan^{2} (y) (csc^{2} (y) - 1) = 3