Is 6cos^2(B)-6sin^2(B)=12cos^2(B)-6 ?

The answer to whether 6cos^2(B)-6sin^2(B)=12cos^2(B)-6 is True

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prove 6cos^2(B)-6sin^2(B)=12cos^2(B)-6

prove

6 cos^{2} (B) - 6 sin^{2} (B) = 12 cos^{2} (B) - 6

True

Solution steps

6 cos^{2} (B) - 6 sin^{2} (B) = 12 cos^{2} (B) - 6

Manipulating left side

6 cos^{2} (B) - 6 sin^{2} (B)

Rewrite using trig identities

6 cos^{2} (B) - 6 sin^{2} (B)

Use the Pythagorean identity:

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

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

= 6 cos^{2} (B) - 6 (1 - cos^{2} (B))

Simplify

6 cos^{2} (B) - 6 (1 - cos^{2} (B)) : 12 cos^{2} (B) - 6

6 cos^{2} (B) - 6 (1 - cos^{2} (B))

Expand

- 6 (1 - cos^{2} (B)) : - 6 + 6 cos^{2} (B)

- 6 (1 - cos^{2} (B))

Apply the distributive law:

a (b - c) = ab - a c

a = - 6, b = 1, c = cos^{2} (B)

= - 6 \cdot 1 - (- 6) cos^{2} (B)

Apply minus-plus rules

- (- a) = a

= - 6 \cdot 1 + 6 cos^{2} (B)

Multiply the numbers:

6 \cdot 1 = 6

= - 6 + 6 cos^{2} (B)

= 6 cos^{2} (B) - 6 + 6 cos^{2} (B)

Simplify

6 cos^{2} (B) - 6 + 6 cos^{2} (B) : 12 cos^{2} (B) - 6

6 cos^{2} (B) - 6 + 6 cos^{2} (B)

Group like terms

= 6 cos^{2} (B) + 6 cos^{2} (B) - 6

Add similar elements:

6 cos^{2} (B) + 6 cos^{2} (B) = 12 cos^{2} (B)

= 12 cos^{2} (B) - 6

= 12 cos^{2} (B) - 6

= 12 cos^{2} (B) - 6

= 12 cos^{2} (B) - 6

We showed that the two sides could take the same form

\Rightarrow True

cos^{3} (x) = \frac{3}{4} cos (x) + \frac{1}{4} cos (3 x)

2 sin (x) + csc (x) = 0

\frac{sec ( θ ) cos ( θ )}{tan ( θ )} = cot (θ)

tan (\frac{π}{4} + x) = \frac{1 + sin ( 2 x )}{cos ( 2 x )}

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

Is 6cos^2(B)-6sin^2(B)=12cos^2(B)-6 ?
The answer to whether 6cos^2(B)-6sin^2(B)=12cos^2(B)-6 is True