For how many values of n with 0<=n<=100 is the graph of f(x) = sin(x + n) identical to the graph of g(x) = cos(x)?
I think that I should use the relationship that cos(x)=sin(90-x) but I'm not sure how.
Does that mean there's 16 values of n?
we want
sin(x+n) = cos(x)
sin(x)cos(n) + cos(x)sin(n) = cos(x)
we need cos(n)=0 and sin(n)=1
This is true only for n = 2k*pi + pi/2
for k=0,1,2,3...
that is pi/2, 5pi/2, 9pi/2, ...
100 * 2/pi = 63.6, so the last entry in the list is 61pi/2 = 95.8
To determine the values of n for which the graphs of f(x) = sin(x + n) and g(x) = cos(x) are identical, we can use the relationship cos(x) = sin(90 - x).
Setting f(x) = g(x), we have:
sin(x + n) = cos(x)
Using the identity cos(x) = sin(90 - x), we can rewrite the equation as:
sin(x + n) = sin(90 - x)
Since the sine function is periodic with a period of 360 degrees (or 2π radians), we know that sin(x) = sin(y) if and only if x = y + 2πk or x = π - y + 2πk, where k is an integer.
In this case, we have:
x + n = 90 - x + 2πk or x + n = π - (90 - x) + 2πk
Simplifying these equations, we get:
2x + n = 90 + 2πk or 2x + n = π + 2πk
Rearranging to isolate x, we have:
x = (90 + 2πk - n) / 2 or x = (π + 2πk - n) / 2
Since x must be between 0 and 100, let's solve for the values of n that satisfy these inequalities.
For 0 ≤ x ≤ 100, we have:
(90 + 2πk - n) / 2 ≤ x ≤ (π + 2πk - n) / 2
Multiplying the inequalities through by 2 to remove the denominators:
90 + 2πk - n ≤ 2x ≤ π + 2πk - n
We can now solve for n by considering different cases. Note that k can be any integer.
Case 1: 90 + 2πk - n ≤ 2x ≤ π + 2πk - n
For the left inequality:
90 + 2πk - n ≤ 2x
Plugging in the maximum value of x (= 100), we have:
90 + 2πk - n ≤ 2(100)
90 + 2πk - n ≤ 200
Rearranging the equation, we get:
n ≥ 2πk - 110
For the right inequality:
2x ≤ π + 2πk - n
Plugging in the minimum value of x (= 0), we have:
2(0) ≤ π + 2πk - n
0 ≤ π + 2πk - n
Rearranging the equation, we get:
n ≤ π + 2πk
Case 2: π + 2πk - n ≤ 2x ≤ 90 + 2πk - n
For the left inequality:
π + 2πk - n ≤ 2x
Plugging in the maximum value of x (= 100), we have:
π + 2πk - n ≤ 2(100)
π + 2πk - n ≤ 200
Rearranging the equation, we get:
n ≥ 2πk - 200 + π
Simplifying, we have:
n ≥ 3π/2 + 2πk - 200
For the right inequality:
2x ≤ 90 + 2πk - n
Plugging in the minimum value of x (= 0), we have:
2(0) ≤ 90 + 2πk - n
0 ≤ 90 + 2πk - n
Rearranging the equation, we get:
n ≤ 90 + 2πk
In summary, the values of n that satisfy the inequalities are:
n ≥ 2πk - 110 and n ≤ π + 2πk for Case 1, or
n ≥ 3π/2 + 2πk - 200 and n ≤ 90 + 2πk for Case 2.
To find the number of values of n that satisfy these inequalities, we can consider the possible values of k that make the expressions inside the inequalities positive.
For Case 1:
2πk - 110 > 0
k > 55π
π + 2πk > 0
k > -π/2
For Case 2:
3π/2 + 2πk - 200 > 0
k > (200 - 3π/2) / (2π)
90 + 2πk > 0
k > -45π
Since k must be an integer, the possible values of k are:
k ≥ 56 for Case 1, or
k ≥ -44 for Case 2.
Therefore, there are 56 - (-44) + 1 = 101 possible values of n that satisfy the given conditions.
In conclusion, there are 101 values of n with 0 ≤ n ≤ 100 for which the graphs of f(x) = sin(x + n) and g(x) = cos(x) are identical.
To find the values of n for which the graph of f(x) = sin(x + n) is identical to the graph of g(x) = cos(x), we can start by using the identity cos(x) = sin(90 - x).
Let's set f(x) equal to g(x) and substitute g(x) with cos(x):
sin(x + n) = cos(x)
Now, we can use the identity cos(x) = sin(90 - x) to rewrite cos(x) as sin(90 - x):
sin(x + n) = sin(90 - x)
Next, we need to consider the range of values for n such that 0 <= n <= 100. Since sin(x) is a periodic function with a period of 2π, we need to find the values of x within this period that satisfy the equation.
To find these values, we equate the arguments of sin(x + n) and sin(90 - x):
x + n = 90 - x
Simplifying the equation, we get:
2x + n = 90
Now, let's consider the range of values for x that satisfy the equation. Since sin(x) repeats every 2π, we want x + n to be within the range of -π/2 and π/2. This means -π/2 ≤ x + n ≤ π/2.
By solving the equation 2x + n = 90 subject to the range of x, we can find the potential solutions for n.
For -π/2 ≤ x ≤ π/2, we have:
-π/2 ≤ 2x + n ≤ π/2
-2x - π/2 ≤ n ≤ -2x + π/2
Now, we need to find the common range for n within the condition 0 ≤ n ≤ 100.
Considering the lower bound:
0 ≤ -2x - π/2
0 ≤ -2x
0 ≥ x
Since x cannot be negative in the given range, we disregard this case.
Now, let's find the upper bound:
-2x + π/2 ≤ 100
π/2 - 100 ≤ -2x
50 - 100/π ≤ x
This means that the upper bound for x is 50 - 100/π. We'll use this value to find the upper bound for n.
-π/2 ≤ 2x + n ≤ π/2
-π/2 - 2(50 - 100/π) ≤ n
-25π - 100/π ≤ n
Therefore, the values of n that satisfy the equation and fall within the range 0 ≤ n ≤ 100 are:
-25π - 100/π ≤ n ≤ 100
You can use a calculator to compute the exact range of values for n in the range 0 ≤ n ≤ 100.