Is sin(3θ)+sin(θ)=2sin(2θ)cos(θ) ?

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

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prove sin(3θ)+sin(θ)=2sin(2θ)cos(θ)

prove

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

True

Solution steps

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

Manipulating left side

sin (3 θ) + sin (θ)

Rewrite using trig identities

sin (3 θ) + sin (θ)

Use the Sum to Product identity:

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

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

Simplify

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

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

\frac{3 θ + θ}{2} = 2 θ

\frac{3 θ + θ}{2}

Add similar elements:

3 θ + θ = 4 θ

= \frac{4 θ}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 θ

= 2 sin (2 θ) cos (\frac{3 θ - θ}{2})

\frac{3 θ - θ}{2} = θ

\frac{3 θ - θ}{2}

Add similar elements:

3 θ - θ = 2 θ

= \frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

= 2 sin (2 θ) cos (θ)

= 2 sin (2 θ) cos (θ)

= 2 sin (2 θ) cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e - cot (x) + sec (x) csc (x) = tan (x)

p ro v e cos (\frac{3 π}{2}) = cos (\frac{π}{2})

p ro v e csc (\frac{π}{2} - x) cos (x) = sec (x) cos (x)

p ro v e sec (A) - cos (A) = tan (A) sin (A)

p ro v e cos^{2} (z) + sin^{2} (z) = 1

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