Is sin(4θ)= 3/8-cos(2θ)+1/8 cos(4θ) ?

The answer to whether sin(4θ)= 3/8-cos(2θ)+1/8 cos(4θ) is False

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prove sin(4θ)= 3/8-cos(2θ)+1/8 cos(4θ)

Solution

prove

sin (4 θ) = \frac{3}{8} - cos (2 θ) + \frac{1}{8} cos (4 θ)

Solution

False

Solution steps

sin (4 θ) = \frac{3}{8} - cos (2 θ) + \frac{1}{8} cos (4 θ)

Show that the two sides are not equal

For

sin (4 θ) = \frac{3}{8} - cos (2 θ) + \frac{1}{8} cos (4 θ)

plug in

θ = 0

sin (4 \cdot 0) = 0

sin (4 \cdot 0)

Simplify

= sin (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

= 0

\frac{3}{8} - cos (2 \cdot 0) + \frac{1}{8} cos (4 \cdot 0) = - \frac{1}{2} (Decimal : - 0.5)

\frac{3}{8} - cos (2 \cdot 0) + \frac{1}{8} cos (4 \cdot 0)

Simplify:

2 \cdot 0 = 0

2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

Simplify:

4 \cdot 0 = 0

4 \cdot 0

Apply rule

0 \cdot a = 0

= 0

\frac{3}{8} - cos (0) + \frac{1}{8} cos (0) = \frac{3 - 7 cos ( 0 )}{8}

\frac{3}{8} - cos (0) + \frac{1}{8} cos (0)

Add similar elements:

- cos (0) + \frac{1}{8} cos (0) = - \frac{7}{8} cos (0)

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

Factor out common term

cos (0)

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

- 1 + \frac{1}{8} = - \frac{7}{8}

- 1 + \frac{1}{8}

Convert element to fraction:

1 = \frac{1 \cdot 8}{8}

= - \frac{1 \cdot 8}{8} + \frac{1}{8}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot 8 + 1}{8}

- 1 \cdot 8 + 1 = - 7

- 1 \cdot 8 + 1

Multiply the numbers:

1 \cdot 8 = 8

= - 8 + 1

Add/Subtract the numbers:

- 8 + 1 = - 7

= - 7

= \frac{- 7}{8}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{7}{8}

= - \frac{7}{8} cos (0)

= \frac{3}{8} - \frac{7}{8} cos (0)

Multiply

\frac{7}{8} cos (0) : \frac{7 cos ( 0 )}{8}

\frac{7}{8} cos (0)

Multiply fractions:

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

= \frac{7 cos ( 0 )}{8}

= \frac{3}{8} - \frac{7 cos ( 0 )}{8}

Apply rule

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

= \frac{3 - 7 cos ( 0 )}{8}

= \frac{3 - 7 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{3 - 7 \cdot 1}{8}

Simplify

\frac{3 - 7 \cdot 1}{8} : - \frac{1}{2}

\frac{3 - 7 \cdot 1}{8}

3 - 7 \cdot 1 = - 4

3 - 7 \cdot 1

Multiply the numbers:

7 \cdot 1 = 7

= 3 - 7

Subtract the numbers:

3 - 7 = - 4

= - 4

= \frac{- 4}{8}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{8}

Cancel the common factor:

4

= - \frac{1}{2}

= - \frac{1}{2}

The two sides are not equal

\Rightarrow False

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Frequently Asked Questions (FAQ)

Is sin(4θ)= 3/8-cos(2θ)+1/8 cos(4θ) ?
The answer to whether sin(4θ)= 3/8-cos(2θ)+1/8 cos(4θ) is False