Is cos(2A)=2cos^2(A)-2 ?

The answer to whether cos(2A)=2cos^2(A)-2 is False

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prove cos(2A)=2cos^2(A)-2

prove

cos (2 A) = 2 cos^{2} (A) - 2

False

Solution steps

cos (2 A) = 2 cos^{2} (A) - 2

Show that the two sides are not equal

For

cos (2 A) = 2 cos^{2} (A) - 2

plug in

A = 0

cos (2 \cdot 0) = 1

cos (2 \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

2 cos^{2} (0) - 2 = 0

2 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

= 2 \cdot 1^{2} - 2

Simplify

= 0

The two sides are not equal

\Rightarrow False

(sec^{2} (x) - 1) cot^{2} (x) = 1

cos (x) = - \frac{1}{2}

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

csc^{2} (x) - 1 = \frac{cos ( x )}{tan ( x ) sin ( x )}

\frac{4 cot ( x )}{sec ( x )} = 4 csc (x) - 4 sin (x)