Is 1+cos(6x)=2cos^2(3x) ?

The answer to whether 1+cos(6x)=2cos^2(3x) is True

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prove 1+cos(6x)=2cos^2(3x)

prove

1 + cos (6 x) = 2 cos^{2} (3 x)

True

Solution steps

1 + cos (6 x) = 2 cos^{2} (3 x)

Manipulating left side

1 + cos (6 x)

Rewrite using trig identities

1 + cos (6 x)

= 1 + cos (2 \cdot 3 x)

Use the Double Angle identity:

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

= 1 + 2 cos^{2} (3 x) - 1

Simplify

1 + 2 cos^{2} (3 x) - 1 : 2 cos^{2} (3 x)

1 + 2 cos^{2} (3 x) - 1

Group like terms

= 2 cos^{2} (3 x) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (3 x)

= 2 cos^{2} (3 x)

= 2 cos^{2} (3 x)

We showed that the two sides could take the same form

\Rightarrow True

csc^{3} (x) = \frac{1}{sin ^{3} ( x )}

\frac{sec ( a )}{cot ( a ) + tan ( a )} = sin (a)

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

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

\frac{1}{2} \frac{tan ( b )}{1 - cos ^{2} ( b )} = csc (2 b)