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

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

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

prove

cos (x + \frac{π}{2}) = - sin (x)

True

Solution steps

cos (x + \frac{π}{2}) = - sin (x)

Manipulating left side

cos (x + \frac{π}{2})

Rewrite using trig identities

cos (x + \frac{π}{2})

Use the Angle Sum identity:

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

= cos (x) cos (\frac{π}{2}) - sin (x) sin (\frac{π}{2})

Simplify

cos (x) cos (\frac{π}{2}) - sin (x) sin (\frac{π}{2}) : - sin (x)

cos (x) cos (\frac{π}{2}) - sin (x) sin (\frac{π}{2})

cos (x) cos (\frac{π}{2}) = 0

cos (x) cos (\frac{π}{2})

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 cos (x)

Apply rule

0 \cdot a = 0

= 0

sin (x) sin (\frac{π}{2}) = sin (x)

sin (x) sin (\frac{π}{2})

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 sin (x)

Multiply:

sin (x) \cdot 1 = sin (x)

= sin (x)

= 0 - sin (x)

0 - sin (x) = - sin (x)

= - sin (x)

= - sin (x)

= - sin (x)

We showed that the two sides could take the same form

\Rightarrow True

csc^{2} (x) tan (x) - tan (x) = cot (x)

sin (\frac{π}{2} - x) = cos (x)

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

cos^{2} (x) = \frac{1 + cos ( 2 x )}{2}

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