Is cos^2(2x)=(1+cos(4x))/2 ?

The answer to whether cos^2(2x)=(1+cos(4x))/2 is True

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prove cos^2(2x)=(1+cos(4x))/2

prove

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

True

Solution steps

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

Manipulating right side

\frac{1 + cos ( 4 x )}{2}

Rewrite using trig identities

\frac{1 + cos ( 4 x )}{2}

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

Use the Double Angle identity:

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

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

Simplify

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

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

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

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

Group like terms

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

1 - 1 = 0

= 2 cos^{2} (2 x)

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

Divide the numbers:

\frac{2}{2} = 1

= cos^{2} (2 x)

= cos^{2} (2 x)

= cos^{2} (2 x)

We showed that the two sides could take the same form

\Rightarrow True

sin^{2} (θ) csc^{2} (θ) - sin^{2} (θ) = cos^{2} (θ)

cos (x + \frac{π}{3}) + sin (x - \frac{π}{6}) = 0

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

\frac{sin ( θ ) + cos ( θ )}{cos ( θ )} = 1 + tan (θ)

\frac{8}{cos ( x ) + 1} + \frac{8}{cos ( x ) - 1} = - 16 csc (x) cot (x)

Is cos^2(2x)=(1+cos(4x))/2 ?
The answer to whether cos^2(2x)=(1+cos(4x))/2 is True