Is 3cos^2(θ)+sin^2(θ)=3-2sin^2(θ) ?

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

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prove 3cos^2(θ)+sin^2(θ)=3-2sin^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

3 cos^{2} (θ) + sin^{2} (θ)

Rewrite using trig identities

3 cos^{2} (θ) + sin^{2} (θ)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

= 3 (1 - sin^{2} (θ)) + sin^{2} (θ)

Simplify

3 (1 - sin^{2} (θ)) + sin^{2} (θ) : 3 - 2 sin^{2} (θ)

3 (1 - sin^{2} (θ)) + sin^{2} (θ)

Expand

3 (1 - sin^{2} (θ)) : 3 - 3 sin^{2} (θ)

3 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (θ)

= 3 \cdot 1 - 3 sin^{2} (θ)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 sin^{2} (θ)

= 3 - 3 sin^{2} (θ) + sin^{2} (θ)

Add similar elements:

- 3 sin^{2} (θ) + sin^{2} (θ) = - 2 sin^{2} (θ)

= 3 - 2 sin^{2} (θ)

= 3 - 2 sin^{2} (θ)

= 3 - 2 sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

sin (\frac{2 π}{3}) = \frac{3}{2}

\frac{1}{2} 2 sin (x) cos (x) = \frac{1}{2} sin (2 x)

1 - tan^{2} (x) = sec^{2} (x)

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