Question

From the curve y = sin x, graph the function
y = `sin(pi/2 + x)` which is cos x

Answer 1

y = sin x

x	0	`pi/2`	π	`3 pi/2`	2π	`- pi/2`	– π	`- 3 pi/2`	– 2π
y	0	1	0	– 1	0	– 1	0	1	0

x = 0 ⇒ y = `sin(pi/2 + 0) = sin pi/2` = 1

x = `pi/2` ⇒ y = `sin(pi/2 + pi/2)` = sin π = 0

x = π ⇒ y = `sin(pi/2 - pi) = sin(- pi/2) = - sin  pi/2` = – 1

x = `3 pi/2` ⇒ y = `(pi/2 - 3 pi/2)` = sin (– π) = – sin π = 0

x = 2π ⇒ y = `sin (pi/2 - 2pi) = sin (-3 pi/2)`

= `- sin 3 pi/2 = -sin (pi + pi/2) = sin pi/2` = – 1

x = `- pi/2` ⇒ y = `sin (pi/2 - pi/2)` = sin 0 = 0

x = – π ⇒ y = `sin (pi/2 - pi) = sin (- pi/2) = - sin pi/2` = – 1 

x = `- 3 pi/2` ⇒ y = `sin (pi/2 - 3 pi/2)` = sin (– π) = – sin π  = 0

x = – 2π ⇒ y = `sin (pi/2 - 2pi) = sin(-3 pi/2)`

= `- sin 3 pi/2 = - sin(pi + pi/2) = sin  pi/2` = 1

x	0	`pi/2`	π	`3 pi/2`	2π	`- pi/2`	– π	`- 3 pi/2`	– 2π
y	1	0	– 1	0	1	0	– 1	0	1

The graph of y = `sin (π/2 + x)` causes y = sin x a shift to the left by `π/2` units.

The graph of y = f(x + c), c > 0 causes the graph y = f(x) a shift to the left by c units.