Is (sin(θ)+sin(3θ))/(2cos(θ))=sin(2θ) ?

The answer to whether (sin(θ)+sin(3θ))/(2cos(θ))=sin(2θ) is True

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prove (sin(θ)+sin(3θ))/(2cos(θ))=sin(2θ)

prove

\frac{sin ( θ ) + sin ( 3 θ )}{2 cos ( θ )} = sin (2 θ)

True

Solution steps

\frac{sin ( θ ) + sin ( 3 θ )}{2 cos ( θ )} = sin (2 θ)

Manipulating left side

\frac{sin ( θ ) + sin ( 3 θ )}{2 cos ( θ )}

Rewrite using trig identities

\frac{sin ( θ ) + sin ( 3 θ )}{2 cos ( θ )}

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 + 3 θ}{2} ) cos ( \frac{θ - 3 θ}{2} )}{2 cos ( θ )}

Simplify

\frac{2 sin ( \frac{θ + 3 θ}{2} ) cos ( \frac{θ - 3 θ}{2} )}{2 cos ( θ )} : sin (2 θ)

\frac{2 sin ( \frac{θ + 3 θ}{2} ) cos ( \frac{θ - 3 θ}{2} )}{2 cos ( θ )}

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

2 sin (\frac{θ + 3 θ}{2}) cos (\frac{θ - 3 θ}{2})

Add similar elements:

θ + 3 θ = 4 θ

= 2 sin (\frac{4 θ}{2}) cos (\frac{θ - 3 θ}{2})

\frac{θ - 3 θ}{2} = - \frac{2 θ}{2}

\frac{θ - 3 θ}{2}

Add similar elements:

θ - 3 θ = - 2 θ

= \frac{- 2 θ}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2 θ}{2}

= 2 sin (\frac{4 θ}{2}) cos (- \frac{2 θ}{2})

= \frac{2 sin ( \frac{4 θ}{2} ) cos ( - \frac{2 θ}{2} )}{2 cos ( θ )}

Divide the numbers:

\frac{2}{2} = 1

= \frac{sin ( \frac{4 θ}{2} ) cos ( - \frac{2 θ}{2} )}{cos ( θ )}

Use the negative angle identity:

cos (- x) = cos (x)

= \frac{cos ( \frac{2 θ}{2} ) sin ( \frac{4 θ}{2} )}{cos ( θ )}

Divide the numbers:

\frac{2}{2} = 1

= \frac{cos ( θ ) sin ( \frac{4 θ}{2} )}{cos ( θ )}

Divide the numbers:

\frac{4}{2} = 2

= \frac{cos ( θ ) sin ( 2 θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= sin (2 θ)

= sin (2 θ)

= sin (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

Is (sin(θ)+sin(3θ))/(2cos(θ))=sin(2θ) ?
The answer to whether (sin(θ)+sin(3θ))/(2cos(θ))=sin(2θ) is True