Is cos(pi+x)=cos(x) ?

The answer to whether cos(pi+x)=cos(x) is False

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prove cos(pi+x)=cos(x)

prove

cos (π + x) = cos (x)

False

Solution steps

cos (π + x) = cos (x)

Show that the two sides are not equal

For

cos (π + x) = cos (x)

plug in

x = 0

cos (π + 0) = - 1

cos (π + 0)

Simplify

= cos (π)

Use the following trivial identity:

cos (π) = (- 1)

cos (π)

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

cos (0) = 1

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

The two sides are not equal

\Rightarrow False

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

(sin^{2} (x) - 1)^{2} = cos (2 x) + sin^{4} (x)

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

csc^{4} (x) - cot^{4} (x) = 2 csc^{2} (x) + 1

(1 - cos (x)) (1 + sec (x)) = sin (x) cos (x)