Is 4cos^2(x)-4sin^2(x)=4-8sin^2(x) ?

The answer to whether 4cos^2(x)-4sin^2(x)=4-8sin^2(x) is True

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prove 4cos^2(x)-4sin^2(x)=4-8sin^2(x)

prove

4 cos^{2} (x) - 4 sin^{2} (x) = 4 - 8 sin^{2} (x)

True

Solution steps

4 cos^{2} (x) - 4 sin^{2} (x) = 4 - 8 sin^{2} (x)

Manipulating left side

4 cos^{2} (x) - 4 sin^{2} (x)

Rewrite using trig identities

4 cos^{2} (x) - 4 sin^{2} (x)

Use the Pythagorean identity:

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

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

= 4 (1 - sin^{2} (x)) - 4 sin^{2} (x)

Simplify

4 (1 - sin^{2} (x)) - 4 sin^{2} (x) : 4 - 8 sin^{2} (x)

4 (1 - sin^{2} (x)) - 4 sin^{2} (x)

Expand

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

= 4 \cdot 1 - 4 sin^{2} (x)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 sin^{2} (x)

= 4 - 4 sin^{2} (x) - 4 sin^{2} (x)

Add similar elements:

- 4 sin^{2} (x) - 4 sin^{2} (x) = - 8 sin^{2} (x)

= 4 - 8 sin^{2} (x)

= 4 - 8 sin^{2} (x)

= 4 - 8 sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{cos ^{2} ( x )}{sin ^{2} ( x )} = cot^{2} (x)

\frac{1 + tan ^{2} ( θ )}{sec ^{2} ( θ )} = 1

cos^{2} (A) + cot^{2} (A) cos^{2} (A) = cot^{2} (A)

cot (9 0^{\circ}) = \frac{1}{tan ( 9 0 ^{\circ} )}

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

Is 4cos^2(x)-4sin^2(x)=4-8sin^2(x) ?
The answer to whether 4cos^2(x)-4sin^2(x)=4-8sin^2(x) is True