What is the value of -cos(pi/2+pi/2) ?

The value of -cos(pi/2+pi/2) is 1

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-cos(pi/2+pi/2)

- cos (\frac{π}{2} + \frac{π}{2})

1

Solution steps

- cos (\frac{π}{2} + \frac{π}{2})

Rewrite using trig identities:

cos (\frac{π}{2} + \frac{π}{2}) = - 1

cos (\frac{π}{2} + \frac{π}{2})

Simplify:

\frac{π}{2} + \frac{π}{2} = π

\frac{π}{2} + \frac{π}{2}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{π + π}{2}

Add similar elements:

π + π = 2 π

= \frac{2 π}{2}

Divide the numbers:

\frac{2}{2} = 1

= π

= 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

= - (- 1)

Simplify

= 1

sin (arccos (\frac{2}{9}))

\frac{3.7}{tan ( 4.6 )}

ln (2 + tan (\frac{π}{16}))

(cos (π))^{2}

arctan (\frac{- \frac{1}{2}}{- \frac{3}{2}})