Question

the equation are of the form y=a sin n(x+b) + c
find the values of a, n, b, c. Hence, write the equation of the function. Attached is the graph

Answer 1

a=2, period = 1 , c= -1

Answer 2

but i dont know how to find b

Answer 3

Notice that \(f(\pi/6)=1\) in other words:\[
2\sin(\pi/6+b) - 1 = 1
\]

Answer 4

This simplifies to \[
\sin(\pi/6+b) = 1
\]

Answer 5

Since \(\sin(\pi/2)=1\) then we could suppose: \[
\pi/2 = \pi/6+b
\]

Answer 6

why did you chose pi/6 ?

Answer 7

so that you get a y value?

Answer 8

Any point would work. \(f(\pi/6)=1\) is the first point we can most obviously see

Answer 9

You can also do \(\pi/2\) since it is clear \(f(\pi/2)=0\)

Answer 10

ok now i understand i have one more question how did you work out sin(π/6+b)=1?

Answer 11

Try to work out \[
2\sin(\pi/2+b)-1 = 0
\]See what happens.

Answer 12

I used algebra.

Answer 13

sin(π/2+b)=1/2

Answer 14

yes

Answer 15

sorry i meant π/2=π/6+b

Answer 16

do both sines on each side cancel each other?

Answer 17

\[
\sin(\pi/6+b) = 1 = \sin(\pi/2)
\]

Answer 18

Technically speaking: \[
\pi/6+b = \pi/2+2\pi n
\]But we just let \(n=0\) to make it simple.

Answer 19

where did the sines go?

Answer 20

do they cancel?

Answer 21

You can do \(\sin^{-1}(\ldots)\) to both sides.

Answer 22

and what happens?

Answer 23

Well \(
\sin^{-1}(\sin(\dots)) = \dots
\)

Answer 24

Typically speaking...

Answer 25

thanks a lot for your help. you explained it very well thank you

Answer 26

We cannot do \(f(x_1)=f(x_2)\implies x_1=x_2\) in general.

Answer 27

However, this is a more special case that allows us to do something along these lines.

Answer 28

cheers