Is sin(θ)cos(φ)=(sin(θ+φ)+sin(θ-φ))/2 ?

The answer to whether sin(θ)cos(φ)=(sin(θ+φ)+sin(θ-φ))/2 is True

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prove sin(θ)cos(φ)=(sin(θ+φ)+sin(θ-φ))/2

prove

sin (θ) cos (φ) = \frac{sin ( θ + φ ) + sin ( θ - φ )}{2}

True

Solution steps

sin (θ) cos (φ) = \frac{sin ( θ + φ ) + sin ( θ - φ )}{2}

Manipulating right side

\frac{sin ( θ + φ ) + sin ( θ - φ )}{2}

Rewrite using trig identities

\frac{sin ( θ + φ ) + sin ( θ - φ )}{2}

Use the Sum to Product identity:

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

= \frac{2 sin ( \frac{θ + φ + θ - φ}{2} ) cos ( \frac{θ + φ - ( θ - φ )}{2} )}{2}

Simplify

\frac{2 sin ( \frac{θ + φ + θ - φ}{2} ) cos ( \frac{θ + φ - ( θ - φ )}{2} )}{2} : sin (θ) cos (φ)

\frac{2 sin ( \frac{θ + φ + θ - φ}{2} ) cos ( \frac{θ + φ - ( θ - φ )}{2} )}{2}

θ + φ + θ - φ = 2 θ

θ + φ + θ - φ

Group like terms

= θ + θ + φ - φ

Add similar elements:

θ + θ = 2 θ

= 2 θ + φ - φ

Add similar elements:

φ - φ = 0

= 2 θ

= \frac{2 cos ( \frac{θ + φ - ( θ - φ )}{2} ) sin ( \frac{2 θ}{2} )}{2}

Divide the numbers:

\frac{2}{2} = 1

= cos (\frac{θ + φ - ( θ - φ )}{2}) sin (\frac{2 θ}{2})

\frac{θ + φ - ( θ - φ )}{2} = φ

\frac{θ + φ - ( θ - φ )}{2}

Expand

θ + φ - (θ - φ) : 2 φ

θ + φ - (θ - φ)

- (θ - φ) : - θ + φ

- (θ - φ)

Distribute parentheses

= - (θ) - (- φ)

Apply minus-plus rules

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

= - θ + φ

= θ + φ - θ + φ

Simplify

θ + φ - θ + φ : 2 φ

θ + φ - θ + φ

Group like terms

= θ - θ + φ + φ

Add similar elements:

θ - θ = 0

= φ + φ

Add similar elements:

φ + φ = 2 φ

= 2 φ

= 2 φ

= \frac{2 φ}{2}

Divide the numbers:

\frac{2}{2} = 1

= φ

= sin (\frac{2 θ}{2}) cos (φ)

Divide the numbers:

\frac{2}{2} = 1

= sin (θ) cos (φ)

= sin (θ) cos (φ)

= sin (θ) cos (φ)

We showed that the two sides could take the same form

\Rightarrow True

Is sin(θ)cos(φ)=(sin(θ+φ)+sin(θ-φ))/2 ?
The answer to whether sin(θ)cos(φ)=(sin(θ+φ)+sin(θ-φ))/2 is True