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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

sin^{2} (0) cos^{2} (0) = 0

sin^{2} (0) cos^{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

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

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

Simplify

= 0

\frac{1}{8} (1 - sin (4 \cdot 0)) = \frac{1}{8} (Decimal : 0.125)

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

Simplify:

4 \cdot 0 = 0

4 \cdot 0

Apply rule

0 \cdot a = 0

= 0

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

\frac{1}{8} (1 - sin (0))

Multiply fractions:

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

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

1 \cdot (1 - sin (0)) = 1 - sin (0)

1 \cdot (1 - sin (0))

Multiply:

1 \cdot (1 - sin (0)) = (1 - sin (0))

= (1 - sin (0))

Remove parentheses:

(a) = a

= 1 - sin (0)

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

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

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 - 0}{8}

Simplify

= \frac{1}{8}

The two sides are not equal

\Rightarrow False

\frac{1 - tan ( x )}{1 - cot ( x )} = - tan (x)

(sin (A) + cos (A))^{2} = 1 + 2 sin (A) cos (A)

\frac{cos ( b )}{sec ( b )} + \frac{sin ( b )}{csc ( b )} = csc^{2} (b) - cot^{2} (b)

2 cos^{2} (\frac{θ}{2}) - cos (θ) - 1 = 0

\frac{1}{csc ( θ ) - cot ( θ )} = csc (θ) + cot (θ)

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