Is 2cos^2(θ/2)-cos(θ)-1=0 ?

The answer to whether 2cos^2(θ/2)-cos(θ)-1=0 is True

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prove 2cos^2(θ/2)-cos(θ)-1=0

prove

2 cos^{2} (\frac{θ}{2}) - cos (θ) - 1 = 0

True

Solution steps

2 cos^{2} (\frac{θ}{2}) - cos (θ) - 1 = 0

Let:

u = \frac{θ}{2}

2 cos^{2} (u) - cos (2 u) - 1 = 0

Prove

2 cos^{2} (u) - cos (2 u) - 1 = 0 :

True

2 cos^{2} (u) - cos (2 u) - 1 = 0

Manipulating left side

2 cos^{2} (u) - cos (2 u) - 1

Rewrite using trig identities

2 cos^{2} (u) - cos (2 u) - 1

Use the Double Angle identity:

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

= - cos (2 u) + cos (2 u)

Add similar elements:

- cos (2 u) + cos (2 u) = 0

= 0

= 0

We showed that the two sides could take the same form

\Rightarrow True

Therefore

2 cos^{2} (\frac{θ}{2}) - cos (θ) - 1 = 0

We showed that the two sides could take the same form

\Rightarrow True

\frac{1}{csc ( θ ) - cot ( θ )} = csc (θ) + cot (θ)

1 + cot^{2} (- x) = csc^{2} (x)

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

5 cos^{2} (x) + 7 sin^{2} (x) - 5 = 2 sin^{2} (x)

sin (θ) - sin (θ - 2 π) = 0

Is 2cos^2(θ/2)-cos(θ)-1=0 ?
The answer to whether 2cos^2(θ/2)-cos(θ)-1=0 is True