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

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

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

prove

cos (t) = sin (t + \frac{π}{2})

True

Solution steps

cos (t) = sin (t + \frac{π}{2})

Manipulating right side

sin (t + \frac{π}{2})

Rewrite using trig identities

sin (t + \frac{π}{2})

Use the Angle Sum identity:

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

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

Simplify

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

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

sin (t) cos (\frac{π}{2}) = 0

sin (t) 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 sin (t)

Apply rule

0 \cdot a = 0

= 0

cos (t) sin (\frac{π}{2}) = cos (t)

cos (t) 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 cos (t)

Multiply:

cos (t) \cdot 1 = cos (t)

= cos (t)

= 0 + cos (t)

0 + cos (t) = cos (t)

= cos (t)

= cos (t)

= cos (t)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e \frac{sin ^{2} ( t )}{tan ( t )} = sin (t) cos (t)

p ro v e - 2 (2 sin (x) cos (x)) = - 2 sin (2 x)

p ro v e (sin (A) - cos (A))^{2} = 1 - sin (2 A)

p ro v e - sec^{2} (x) = - \frac{1}{cos ^{2} ( x )}

p ro v e sin (2 x) = 3 sin (x)