Is sin^2(y)cos^2(y)=(1-cos(4x))/8 ?

The answer to whether sin^2(y)cos^2(y)=(1-cos(4x))/8 is False

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prove sin^2(y)cos^2(y)=(1-cos(4x))/8

prove

sin^{2} (y) cos^{2} (y) = \frac{1 - cos ( 4 x )}{8}

False

Solution steps

sin^{2} (y) cos^{2} (y) = \frac{1 - cos ( 4 x )}{8}

Show that the two sides are not equal

For

sin^{2} (y) cos^{2} (y) = \frac{1 - cos ( 4 x )}{8}

plug in

x = 0

For

sin^{2} (y) cos^{2} (y) = \frac{1 - cos ( 4 x )}{8}

plug in

y = 1

sin^{2} (1) cos^{2} (1) = 0.20670 \dots

sin^{2} (1) cos^{2} (1)

Refine to a decimal form

= 0.20670 \dots

\frac{1 - cos ( 4 \cdot 0 )}{8} = 0

\frac{1 - cos ( 4 \cdot 0 )}{8}

Simplify

= \frac{1 - cos ( 0 )}{8}

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 - 1}{8}

Simplify

= 0

The two sides are not equal

\Rightarrow False

cos (θ) tan (θ) sin (θ) = sin^{2} (θ)

cos (\frac{3 x}{5}) = - sin (\frac{3 x}{5} - \frac{π}{2})

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

\frac{( cos ( a - b ) )}{sin ( a ) sin ( b )} = cot (a) cot (b) + 1

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

Is sin^2(y)cos^2(y)=(1-cos(4x))/8 ?
The answer to whether sin^2(y)cos^2(y)=(1-cos(4x))/8 is False