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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

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

1 - (- cos (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

= 1 - (- 1)^{2}

Simplify

1 - (- 1)^{2} : 0

1 - (- 1)^{2}

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

= 1 - 1

Subtract the numbers:

1 - 1 = 0

= 0

= 0

1 + cos^{2} (0) = 2

1 + 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

= 1 + 1^{2}

Simplify

= 2

The two sides are not equal

\Rightarrow False

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