Is (sin(2θ)+sin(θ))/(cos(2θ)+cos(θ)+1)=tan(θ) ?

The answer to whether (sin(2θ)+sin(θ))/(cos(2θ)+cos(θ)+1)=tan(θ) is True

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prove (sin(2θ)+sin(θ))/(cos(2θ)+cos(θ)+1)=tan(θ)

prove

\frac{sin ( 2 θ ) + sin ( θ )}{cos ( 2 θ ) + cos ( θ ) + 1} = tan (θ)

True

Solution steps

\frac{sin ( 2 θ ) + sin ( θ )}{cos ( 2 θ ) + cos ( θ ) + 1} = tan (θ)

Manipulating left side

\frac{sin ( 2 θ ) + sin ( θ )}{cos ( 2 θ ) + cos ( θ ) + 1}

Rewrite using trig identities

\frac{sin ( 2 θ ) + sin ( θ )}{cos ( 2 θ ) + cos ( θ ) + 1}

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

= \frac{sin ( 2 θ ) + sin ( θ )}{2 cos ^{2} ( θ ) - 1 + cos ( θ ) + 1}

2 cos^{2} (θ) - 1 + cos (θ) + 1 = 2 cos^{2} (θ) + cos (θ)

2 cos^{2} (θ) - 1 + cos (θ) + 1

Group like terms

= 2 cos^{2} (θ) + cos (θ) - 1 + 1

- 1 + 1 = 0

= 2 cos^{2} (θ) + cos (θ)

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

Use the Double Angle identity:

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

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

Simplify

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

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

Factor out common term

sin (θ)

= \frac{sin ( θ ) ( 2 cos ( θ ) + 1 )}{2 cos ^{2} ( θ ) + cos ( θ )}

Factor

2 cos^{2} (θ) + cos (θ) : cos (θ) (2 cos (θ) + 1)

2 cos^{2} (θ) + cos (θ)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

cos^{2} (θ) = cos (θ) cos (θ)

= 2 cos (θ) cos (θ) + cos (θ)

Factor out common term

cos (θ)

= cos (θ) (2 cos (θ) + 1)

= \frac{sin ( θ ) ( 2 cos ( θ ) + 1 )}{cos ( θ ) ( 2 cos ( θ ) + 1 )}

Cancel the common factor:

2 cos (θ) + 1

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

= \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

= tan (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Is (sin(2θ)+sin(θ))/(cos(2θ)+cos(θ)+1)=tan(θ) ?
The answer to whether (sin(2θ)+sin(θ))/(cos(2θ)+cos(θ)+1)=tan(θ) is True