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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

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

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

Refine to a decimal form

= 0

- 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

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

\frac{1}{cos ( x ) + 1} + \frac{1}{tan ( x ) + 1} = 1

\frac{1}{csc ( x ) + 1} + \frac{1}{csc ( x ) - 1} = 2 sec (x) tan (x)

25 cos^{2} (x) = 25 - 25 sin^{2} (x)

\frac{1}{1 - sin ( θ )} + \frac{1}{1 + sin ( θ )} = 4 tan (θ)

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