Is 8sin^4(θ)=cos(4θ)+4cos(2θ)-3 ?

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

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prove 8sin^4(θ)=cos(4θ)+4cos(2θ)-3

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

θ = 0

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

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

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

= 2

The two sides are not equal

\Rightarrow False

sec^{2} (a) csc^{2} (a) = sec^{2} (a) + csc^{2} (a)

cot (x) - tan (x) = 2 tan (2 x)

\frac{csc ( θ ) tan ( θ )}{1 + tan ^{2} ( θ )} = cos (θ)

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

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

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