Is (-cos^2(2θ))/2 =(-1-cos(4θ))/8 ?

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

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prove (-cos^2(2θ))/2 =(-1-cos(4θ))/8

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

θ = 0

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

\frac{- cos ^{2} ( 2 \cdot 0 )}{2}

Simplify

= - \frac{cos ^{2} ( 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}}{2}

Simplify

= - \frac{1}{2}

\frac{- 1 - cos ( 4 \cdot 0 )}{8} = - \frac{1}{4} (Decimal : - 0.25)

\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

\frac{- 1 - 1}{8} : - \frac{1}{4}

\frac{- 1 - 1}{8}

Subtract the numbers:

- 1 - 1 = - 2

= \frac{- 2}{8}

Apply the fraction rule:

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

= - \frac{2}{8}

Cancel the common factor:

2

= - \frac{1}{4}

= - \frac{1}{4}

The two sides are not equal

\Rightarrow False

p ro v e \frac{1 - cos ( x )}{- 1 + sec ( x )} = cos (x)

p ro v e 2 sin (y) cos (y) sec (2 y) = tan (2 y)

p ro v e tan (\frac{3 π}{2} - x) = - tan (x)

p ro v e - 3 (0) cos (3 t) + 3 (1) sin (3 t) = 3 sin (3 t)

p ro v e (1 + cos (x)) (1 - cos (- x)) = sin^{2} (x)

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