Is cos^2(u)= 1/2+1/2 cos(2u) ?

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

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prove cos^2(u)= 1/2+1/2 cos(2u)

prove

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

True

Solution steps

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

Manipulating right side

\frac{1}{2} + \frac{1}{2} cos (2 u)

Factor

\frac{1}{2} + cos (2 u) \frac{1}{2} : \frac{1}{2} (1 + cos (2 u))

\frac{1}{2} + cos (2 u) \frac{1}{2}

Factor out common term

\frac{1}{2}

= \frac{1}{2} (1 + cos (2 u))

= (1 + cos (2 u)) \frac{1}{2}

Rewrite using trig identities

(1 + cos (2 u)) \frac{1}{2}

Use the Double Angle identity:

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

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

Simplify

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

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

1 \cdot (1 + 2 cos^{2} (u) - 1)

Simplify

1 + 2 cos^{2} (u) - 1 : 2 cos^{2} (u)

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

Group like terms

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

1 - 1 = 0

= 2 cos^{2} (u)

= 1 \cdot 2 cos^{2} (u)

Multiply the numbers:

1 \cdot 2 = 2

= 2 cos^{2} (u)

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

Divide the numbers:

\frac{2}{2} = 1

= cos^{2} (u)

= cos^{2} (u)

= cos^{2} (u)

We showed that the two sides could take the same form

\Rightarrow True

2 sin (- y) cos (x) = 2 sin (- y) cos (x)

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

csc (x^{2}) sec (x^{2}) = csc (x^{2}) + sec (x^{2})

tan (\frac{θ}{2}) = csc^{2} (θ) - cot^{2} (θ)

4 + 4 cos (2 x) = 8 cos^{2} (x)

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