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.
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)
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
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
Just wanted to express my gratitude. Thankyou very much for your help.
No problem! I'm here to help. Let's break down each part of the question.
Part 1: Writing f(x) in the form Asin(x+a)
To write f(x) in the form Asin(x+a), we need to rewrite the expression so that it only involves sine. Let's start by expanding the expression.
f(x) = 2sinx - cosx
Next, we can express cos(x) using the identity: cos(x) = sin(x + π/2).
f(x) = 2sinx - sin(x + π/2)
Now, using the identity sin(A - B) = sin(A)cos(B) - cos(A)sin(B), we can rewrite f(x).
f(x) = 2sinx - sin(x)cos(π/2) - cos(x)sin(π/2)
= 2sinx - sin(x) * 0 - cos(x) * 1
= 2sinx - cos(x)
Therefore, f(x) in the form Asin(x+a) is f(x) = 2sin(x + π/2).
Part 2: Finding the maximum value of 2sinx - cosx
To find the maximum value, we need to identify the amplitude and phase shift of the function 2sinx - cosx.
The amplitude of a sine function is the coefficient of the sine term, which in this case is 2. So, the amplitude is 2.
The phase shift of a sine function is given by the constant in the argument of the sine function, divided by the coefficient of x. In this case, the phase shift is π/2 divided by 1, which is just π/2.
Since the amplitude is positive, and the phase shift is positive, the maximum value of the function occurs when the argument of the sine function is π/2.
So, the maximum value of 2sinx - cosx is 2sin(π/2) - cos(π/2) = 2(1) - 0 = 2.
Part 3: Finding the lowest positive value of p such that f(x) = f(x + p)
To find the lowest positive value of p such that f(x) = f(x + p), we can compare the expressions for f(x) and f(x + p).
f(x) = f(x + p)
2sinx - cosx = 2sin(x + p) - cos(x + p)
Let's expand the expressions.
2sinx - cosx = 2[ sinx*cosp + sinp*cosx ] - [ cosx*cosp - sinx*sinp ]
Now, let's compare the coefficients of sinx and cosx on both sides of the equation.
Coefficient of sinx on the left side: 0
Coefficient of sinx on the right side: 2*sinp
Coefficient of cosx on the left side: -1
Coefficient of cosx on the right side: 2*cosp
For the equation to be true, the coefficients on both sides must be equal. Therefore, we have the following system of equations:
0 = 2*sinp (Equation 1)
-1 = 2*cosp (Equation 2)
Solving Equation 1, we find sinp = 0. Since p is the lowest positive value, this occurs when p = π.
Now, let's substitute p = π into Equation 2.
-1 = 2*cos(π)
-1 = 2*(-1)
-1 = -2
The equation is not true for p = π. Therefore, there is no positive value of p for which f(x) = f(x + p).
I hope this helps! Let me know if you have any further questions.