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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= 4 (1 - cos^{2} (x)) + 2 cos^{2} (x)

Simplify

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

4 (1 - cos^{2} (x)) + 2 cos^{2} (x)

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

4 \cdot 1 = 4

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

= 4 - 4 cos^{2} (x) + 2 cos^{2} (x)

Add similar elements:

- 4 cos^{2} (x) + 2 cos^{2} (x) = - 2 cos^{2} (x)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

2 cos (4 θ) + 5 = 16 sin^{2} (θ)

(1 + cos (x)) (csc (x) - cot (x)) = sin (x)

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

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

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