Is 3sin^2(θ)+4cos^2(θ)=3+cos^2(θ) ?

The answer to whether 3sin^2(θ)+4cos^2(θ)=3+cos^2(θ) is True

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prove 3sin^2(θ)+4cos^2(θ)=3+cos^2(θ)

prove

3 sin^{2} (θ) + 4 cos^{2} (θ) = 3 + cos^{2} (θ)

True

Solution steps

3 sin^{2} (θ) + 4 cos^{2} (θ) = 3 + cos^{2} (θ)

Manipulating left side

3 sin^{2} (θ) + 4 cos^{2} (θ)

Rewrite using trig identities

3 sin^{2} (θ) + 4 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= 3 (1 - cos^{2} (θ)) + 4 cos^{2} (θ)

Simplify

3 (1 - cos^{2} (θ)) + 4 cos^{2} (θ) : 3 + cos^{2} (θ)

3 (1 - cos^{2} (θ)) + 4 cos^{2} (θ)

Expand

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

3 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = cos^{2} (θ)

= 3 \cdot 1 - 3 cos^{2} (θ)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 cos^{2} (θ)

= 3 - 3 cos^{2} (θ) + 4 cos^{2} (θ)

Add similar elements:

- 3 cos^{2} (θ) + 4 cos^{2} (θ) = cos^{2} (θ)

= 3 + cos^{2} (θ)

= 3 + cos^{2} (θ)

= 3 + cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

(1 + cot^{2} (θ)) sin^{2} (θ) = 1

1 - tan^{2} (x) = \frac{cos ( 2 x )}{cos ^{2} ( x )}

sin^{2} (- θ) + cos^{2} (- θ) = 1

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

\frac{cos ( θ )}{sin ( θ ) cot ( θ )} = 1

Is 3sin^2(θ)+4cos^2(θ)=3+cos^2(θ) ?
The answer to whether 3sin^2(θ)+4cos^2(θ)=3+cos^2(θ) is True