Is cos(z)=sin(pi/2+z) ?

The answer to whether cos(z)=sin(pi/2+z) is True

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prove cos(z)=sin(pi/2+z)

prove

cos (z) = sin (\frac{π}{2} + z)

True

Solution steps

cos (z) = sin (\frac{π}{2} + z)

Manipulating right side

sin (\frac{π}{2} + z)

Rewrite using trig identities

sin (\frac{π}{2} + z)

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (\frac{π}{2}) cos (z) + cos (\frac{π}{2}) sin (z)

Simplify

sin (\frac{π}{2}) cos (z) + cos (\frac{π}{2}) sin (z) : cos (z)

sin (\frac{π}{2}) cos (z) + cos (\frac{π}{2}) sin (z)

sin (\frac{π}{2}) cos (z) = cos (z)

sin (\frac{π}{2}) cos (z)

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

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 \cdot cos (z)

Multiply:

1 \cdot cos (z) = cos (z)

= cos (z)

cos (\frac{π}{2}) sin (z) = 0

cos (\frac{π}{2}) sin (z)

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 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}

= 0

= 0 \cdot sin (z)

Apply rule

0 \cdot a = 0

= 0

= cos (z) + 0

cos (z) + 0 = cos (z)

= cos (z)

= cos (z)

= cos (z)

We showed that the two sides could take the same form

\Rightarrow True