Question

Given that f(x) = 2sinx - cosx

- Write f(x) in the form Asin(x+a)

- Find the maximum value of 2sinx - cosx

- Find the lowest positive value of p such that f(x) = f(x + p)

Thankyou. All help is appreciated. I'm really struggling on this assignment.

Answer 1

how about expanding Asin(x+a)
= Asinxcosa + Acosxsina

so Asinxcosa + Acosxsina = 2sinx - cosx

This must now be an identity, that is, it must be true for all values of x
(I will work in degrees, if you need your answers in radians either convert these values or repeat the following steps with your calculator set to radians)

let x = 0
Asin0cosa + Acos0sina = 2sin0 - cos0
0 + Asina = 0 - 1
Asina = -1 (#1)

let x=90º
then Asin90cosa + Acos90sina = 2sin90 - cos90
Acosa + 0 = 2 - 0
Acosa = 2 (#2)

divide equation #1 by #2 to get
sina/cosa = -1/2
tana = -1/2
so a = 153.4º or a = -26.6º,...there are more

using a=153.4 back in #1 gives us A=-2.236
using a=-26.6 back in #1 gives us
A=+2.236

so we could write
f(x) = -2.236sin(x+153.4º) or
f(x) = 2.236sin(x-26.6º)

Answer 2

We could just trust my answer to the first part, and simply say that since │A│ is the amplitude of our function we found, so the maximum value must be 2.236

or

we could Calculus.

the f'(x) = 2cosx + sinx
= 0 for a max/min of f(x)

sinx = -2cosx
sinx/cosx = -2
tanx = -2
x = 116.4º or x=296.6º

going back to our function
f(116.4) = 2.236 and f(296.6)=-2.236

confirming our answer to the first part.

so the max value of f(x) is 2.236

Answer 3

recall from above that we concluded that

f(x) = -2.236sin(x+153.4º) or f(x) = 2.236sin(x-26.6º)

the period of either one is 360º, so a phase shift of 360º either to the left or right would result in the same graph

But the wanted the smallest possible value in (x+p)

so if we added 360 to the existing phase shifts, we would only get a value larger than what we already have in

f(x) = -2.236sin(x+153.4º) so the smallest value of p is 153.4º

Answer 4

Just wanted to express my gratitude. Thankyou very much for your help.