Is 3sin^2(2x) 1/2 =(3cos(4x))/4 ?

The answer to whether 3sin^2(2x) 1/2 =(3cos(4x))/4 is False

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prove 3sin^2(2x) 1/2 =(3cos(4x))/4

prove

3 sin^{2} (2 x) \frac{1}{2} = \frac{3 cos ( 4 x )}{4}

False

Solution steps

3 sin^{2} (2 x) \frac{1}{2} = \frac{3 cos ( 4 x )}{4}

Show that the two sides are not equal

For

3 sin^{2} (2 x) \frac{1}{2} = \frac{3 cos ( 4 x )}{4}

plug in

x = 0

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

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

Simplify:

2 \cdot 0 = 0

2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

3 sin^{2} (0) \frac{1}{2} = \frac{3 sin ^{2} ( 0 )}{2}

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

Multiply fractions:

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

= \frac{1 \cdot 3 sin ^{2} ( 0 )}{2}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3 sin ^{2} ( 0 )}{2}

= \frac{3 sin ^{2} ( 0 )}{2}

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{3 \cdot 0 ^{2}}{2}

Simplify

= 0

\frac{3 cos ( 4 \cdot 0 )}{4} = \frac{3}{4} (Decimal : 0.75)

\frac{3 cos ( 4 \cdot 0 )}{4}

Simplify

= \frac{3 cos ( 0 )}{4}

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{3 \cdot 1}{4}

Simplify

= \frac{3}{4}

The two sides are not equal

\Rightarrow False

sin (3 x) = 3 sin (x) cos^{2} (x) - sin^{2} (x)

cos (θ) - cos^{3} (θ) = sin^{2} (cos (θ))

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

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

cos (30 0^{\circ}) = 1 - 2 sin^{2} (15 0^{\circ})

Is 3sin^2(2x) 1/2 =(3cos(4x))/4 ?
The answer to whether 3sin^2(2x) 1/2 =(3cos(4x))/4 is False