Is cos(2θ)=2cos^2(θ)-1 ?

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

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prove cos(2θ)=2cos^2(θ)-1

prove

cos (2 θ) = 2 cos^{2} (θ) - 1

True

Solution steps

cos (2 θ) = 2 cos^{2} (θ) - 1

Manipulating left side

cos (2 θ)

Rewrite using trig identities

cos (2 θ)

Rewrite as

= cos (θ + θ)

Use the Angle Sum identity:

cos (s + s) = cos (s) cos (s) - sin (s) sin (s)

= cos (θ) cos (θ) - sin (θ) sin (θ)

Simplify

cos (θ) cos (θ) - sin (θ) sin (θ) : cos^{2} (θ) - sin^{2} (θ)

cos (θ) cos (θ) - sin (θ) sin (θ)

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= cos^{2} (θ) - sin^{2} (θ)

= cos^{2} (θ) - sin^{2} (θ)

= cos^{2} (θ) - sin^{2} (θ)

Use the Pythagorean identity:

cos^{2} (θ) + sin^{2} (θ) = 1

sin^{2} (θ) = 1 - cos^{2} (θ)

= cos^{2} (θ) - (1 - cos^{2} (θ))

Simplify

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

cos^{2} (θ) - (1 - cos^{2} (θ))

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

- (1 - cos^{2} (θ))

Distribute parentheses

= - 1 - (- cos^{2} (θ))

Apply minus-plus rules

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

= - 1 + cos^{2} (θ)

= cos^{2} (θ) - 1 + cos^{2} (θ)

Simplify

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

cos^{2} (θ) - 1 + cos^{2} (θ)

Group like terms

= cos^{2} (θ) + cos^{2} (θ) - 1

Add similar elements:

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

= 2 cos^{2} (θ) - 1

= 2 cos^{2} (θ) - 1

= 2 cos^{2} (θ) - 1

We showed that the two sides could take the same form

\Rightarrow True

sin (\frac{7 π}{8})

sin (\frac{4 π}{3})

sin^{2} (x) = 3 cos^{2} (x)

sin (- π)

cos (5)