Is cos(4θ)+4cos(2θ)+3=8sin^4(θ) ?

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

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prove cos(4θ)+4cos(2θ)+3=8sin^4(θ)

prove

cos (4 θ) + 4 cos (2 θ) + 3 = 8 sin^{4} (θ)

False

Solution steps

cos (4 θ) + 4 cos (2 θ) + 3 = 8 sin^{4} (θ)

Show that the two sides are not equal

For

cos (4 θ) + 4 cos (2 θ) + 3 = 8 sin^{4} (θ)

plug in

θ = 0

cos (4 \cdot 0) + 4 cos (2 \cdot 0) + 3 = 8

cos (4 \cdot 0) + 4 cos (2 \cdot 0) + 3

Simplify

= 5 cos (0) + 3

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

= 5 \cdot 1 + 3

Simplify

= 8

8 sin^{4} (0) = 0

8 sin^{4} (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

= 8 \cdot 0^{4}

Simplify

= 0

The two sides are not equal

\Rightarrow False

2 \cdot sin (x) = sin (2 x)

cot (\frac{π}{2} - θ) = \frac{sin ( θ )}{cos ( θ )}

\frac{sec ^{2} ( x )}{tan ( x )} = tan (x) + cot (x)

tan^{2} (A) (csc^{2} (A) - 1) = 1

cos^{3} (θ) = - 1

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