Is 8sin^2(θ)+9cos^2(θ)=8+cos^2(θ) ?

The answer to whether 8sin^2(θ)+9cos^2(θ)=8+cos^2(θ) is True

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prove 8sin^2(θ)+9cos^2(θ)=8+cos^2(θ)

prove

8 sin^{2} (θ) + 9 cos^{2} (θ) = 8 + cos^{2} (θ)

True

Solution steps

8 sin^{2} (θ) + 9 cos^{2} (θ) = 8 + cos^{2} (θ)

Manipulating left side

8 sin^{2} (θ) + 9 cos^{2} (θ)

Rewrite using trig identities

8 sin^{2} (θ) + 9 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= 8 (1 - cos^{2} (θ)) + 9 cos^{2} (θ)

Simplify

8 (1 - cos^{2} (θ)) + 9 cos^{2} (θ) : 8 + cos^{2} (θ)

8 (1 - cos^{2} (θ)) + 9 cos^{2} (θ)

Expand

8 (1 - cos^{2} (θ)) : 8 - 8 cos^{2} (θ)

8 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 8, b = 1, c = cos^{2} (θ)

= 8 \cdot 1 - 8 cos^{2} (θ)

Multiply the numbers:

8 \cdot 1 = 8

= 8 - 8 cos^{2} (θ)

= 8 - 8 cos^{2} (θ) + 9 cos^{2} (θ)

Add similar elements:

- 8 cos^{2} (θ) + 9 cos^{2} (θ) = cos^{2} (θ)

= 8 + cos^{2} (θ)

= 8 + cos^{2} (θ)

= 8 + cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e sec^{2} (x) + tan^{2} (x) = 1 + 2 tan^{2} (x)

p ro v e cot (x + π) = cot (x)

p ro v e tan (x) (2 - cos (x))^{2} = 0

p ro v e 1 - 2 sin^{2} (x) = - 1 + 2 cos^{2} (x)

p ro v e \frac{1}{sin ( x ) cos ( x )} = 2 csc (2 x)

Is 8sin^2(θ)+9cos^2(θ)=8+cos^2(θ) ?
The answer to whether 8sin^2(θ)+9cos^2(θ)=8+cos^2(θ) is True