Is sin(pi/2+x)=-cos(x) ?

The answer to whether sin(pi/2+x)=-cos(x) is False

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prove sin(pi/2+x)=-cos(x)

prove

sin (\frac{π}{2} + x) = - cos (x)

False

Solution steps

sin (\frac{π}{2} + x) = - cos (x)

Show that the two sides are not equal

For

sin (\frac{π}{2} + x) = - cos (x)

plug in

x = 0

sin (\frac{π}{2} + 0) = 1

sin (\frac{π}{2} + 0)

Simplify

= sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (\frac{π}{2})

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{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

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

2 sin (π) = 5 sin (π)

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

\frac{1}{cot ( u )} = \frac{cot ( u )}{1}

sin^{2} (x) \cdot sec (x) = tan^{2} (x)