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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

Multiply the numbers:

4 \cdot 1 = 4

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

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

Add similar elements:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

sec^{2} (x) + 1 = \frac{1 + cos ^{2} ( x )}{cos ^{2} ( x )}

cos (4 A) = 8 cos^{4} (A) - 8 cos^{2} (A) + 1

cos (4 B) = 1 - 8 sin^{2} (B) + 8 sin^{4} (B)

tan (θ) sin (θ) cos (θ) = sin^{2} (θ)

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