Is sin(x/2)*cos(x/2)= 1/2 sin(x) ?

The answer to whether sin(x/2)*cos(x/2)= 1/2 sin(x) is True

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prove sin(x/2)*cos(x/2)= 1/2 sin(x)

prove

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

True

Solution steps

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

Let:

u = \frac{x}{2}

sin (u) cos (u) = \frac{1}{2} sin (2 u)

Prove

sin (u) cos (u) = \frac{1}{2} sin (2 u) :

True

sin (u) cos (u) = \frac{1}{2} sin (2 u)

Manipulating right side

\frac{1}{2} sin (2 u)

Rewrite using trig identities

\frac{1}{2} sin (2 u)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= \frac{1}{2} \cdot 2 sin (u) cos (u)

Simplify

\frac{1}{2} \cdot 2 sin (u) cos (u) : sin (u) cos (u)

\frac{1}{2} \cdot 2 sin (u) cos (u)

Multiply fractions:

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

= \frac{1 \cdot 2}{2} sin (u) cos (u)

Cancel the common factor:

2

= sin (u) cos (u) \cdot 1

Multiply:

cos (u) \cdot 1 = cos (u)

= sin (u) cos (u)

= sin (u) cos (u)

= sin (u) cos (u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

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

We showed that the two sides could take the same form

\Rightarrow True

2 cos (4 θ) + 5 = 16 sin^{2} (θ)

(1 + cos (x)) (csc (x) - cot (x)) = sin (x)

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

\frac{tan ^{2} ( θ )}{sin ( θ )} = sec (θ) tan (θ)

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

Is sin(x/2)*cos(x/2)= 1/2 sin(x) ?
The answer to whether sin(x/2)*cos(x/2)= 1/2 sin(x) is True