Is (1/2 sin^2(θ))=(1/2 cos(2θ)) ?

The answer to whether (1/2 sin^2(θ))=(1/2 cos(2θ)) is False

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prove (1/2 sin^2(θ))=(1/2 cos(2θ))

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

θ = 0

\frac{1}{2} sin^{2} (0) = 0

\frac{1}{2} sin^{2} (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= \frac{1}{2} \cdot 0^{2}

Simplify

= 0

\frac{1}{2} cos (2 \cdot 0) = \frac{1}{2} (Decimal : 0.5)

\frac{1}{2} cos (2 \cdot 0)

Simplify:

2 \cdot 0 = 0

2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

\frac{1}{2} cos (0) = \frac{cos ( 0 )}{2}

\frac{1}{2} cos (0)

Multiply fractions:

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

= \frac{1 \cdot cos ( 0 )}{2}

Multiply:

1 \cdot cos (0) = cos (0)

= \frac{cos ( 0 )}{2}

= \frac{cos ( 0 )}{2}

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= \frac{1}{2}

The two sides are not equal

\Rightarrow False

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

sin (x - π) = sin (x) - sin (π)

cos (- \frac{2}{3} π) = cos (\frac{4 π}{3})

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

6 cos^{2} (A) - 5 = 1 - 6 sin^{2} (A)

Is (1/2 sin^2(θ))=(1/2 cos(2θ)) ?
The answer to whether (1/2 sin^2(θ))=(1/2 cos(2θ)) is False