Is cos(3θ)=4cos^3(θ)-cos(θ) ?

The answer to whether cos(3θ)=4cos^3(θ)-cos(θ) is False

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prove cos(3θ)=4cos^3(θ)-cos(θ)

prove

cos (3 θ) = 4 cos^{3} (θ) - cos (θ)

False

Solution steps

cos (3 θ) = 4 cos^{3} (θ) - cos (θ)

Show that the two sides are not equal

For

cos (3 θ) = 4 cos^{3} (θ) - cos (θ)

plug in

θ = 0

cos (3 \cdot 0) = 1

cos (3 \cdot 0)

Simplify

= cos (0)

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

= 1

4 cos^{3} (0) - cos (0) = 3

4 cos^{3} (0) - cos (0)

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

= 4 \cdot 1^{3} - 1

Simplify

= 3

The two sides are not equal

\Rightarrow False

cos^{2} (b) + cos^{2} (\frac{π}{2} - b) = 1

sec (x) - (sin (x)) (tan (x)) = cos (x)

\frac{1}{tan ( θ )} + tan (θ) = sec (θ) csc (θ)

sec (x) - \frac{1}{sec ( x )} = sin (x) tan (x)

\frac{1 + csc ( b )}{cot ( b ) + cos ( b )} = sec (b)

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