Is 1(-cos(2x))=2cos^2(x) ?

The answer to whether 1(-cos(2x))=2cos^2(x) is False

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prove 1(-cos(2x))=2cos^2(x)

prove

1 (- cos (2 x)) = 2 cos^{2} (x)

False

Solution steps

1 \cdot (- cos (2 x)) = 2 cos^{2} (x)

Show that the two sides are not equal

For

1 (- cos (2 x)) = 2 cos^{2} (x)

plug in

x = 0

1 \cdot (- cos (2 \cdot 0)) = - 1

1 \cdot (- 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

2 cos^{2} (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

= 2 \cdot 1^{2}

Simplify

= 2

The two sides are not equal

\Rightarrow False

- tan^{2} (x) = (tan^{2} (x) + 1) (cos^{2} (x) - 1)

csc (x) tan (x) cos^{2} (x) = cos (x)

sin (2 x) cot (x) = 2 cos^{2} (x)

sin (a) sin (b) = sin (a + b)

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