Please help me understand how to solve this questions. I need step by step solution and answer because I find it difficult to understand when someone on here writes the answer instead of showing step by step work on how he/she got the numbers to get the answers. I know I'm asking for too much, but I want to learn it. I want to understand it so when I take a test in the future, I won't find it challenging. THANK YOU SO MUCH!.
1) Determine the quarter points of
y=−cos(2x−π)
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2) Find the quarter points of y=-3sin(-x+π/2)
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3) Find the quarter points of y=3cos(4x+π)+1
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4) Find the quarter points of
y=2-3cos(4x-π)
(Simplify your answers. Type ORDERED PAIRS. Type exact answers, using π as needed. Use integers or fractions for any numbers in the expressions)
what you really need to know is that the period of cos(kx) is 2π/k.
Then just divide that length into 4 intervals, and evaluate y at each dividing point.
For example,
3cos(4x+π)+1 has period 2π/4 = π/2.
Thus, each quarter has length π/8.
So, the points are
(0,-2)
(π/8,0)
(π/4,4)
(3π/8,0)
(π/2,-2)
Of course, you might want to shift that so it starts at its maximum, the way cos(x) does. In that case, note that
cos(4x+π) = cos(4(x + π/4))
So, you need to shift left by π/4, so the quarter points would be
(-π/4,4)
(-π/8,0)
(0,-2)
(π/8,0)
(π/4,4)
Sure! I'd be happy to help you understand how to solve these problems step by step.
1) Determine the quarter points of y = -cos(2x-π):
To find the quarter points, we need to find the x-values where the function crosses the x-axis. These points are also known as the x-intercepts.
First, set y = 0 and solve for x:
0 = -cos(2x-π)
Since cosine of an angle equals zero when the angle is π/2 or 3π/2, we can set 2x-π equal to these values and solve for x:
2x-π = π/2 or 2x-π = 3π/2
Solving the first equation, we have:
2x = π/2 + π
2x = 3π/2
x = (3π/2) / 2
x = 3π/4
Solving the second equation, we have:
2x = π/2 + 2π
2x = 5π/2
x = (5π/2) / 2
x = 5π/4
Therefore, the quarter points of y = -cos(2x-π) are the ordered pairs (3π/4, 0) and (5π/4, 0).
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2) Find the quarter points of y = -3sin(-x+π/2):
Again, we need to find the x-values where the function crosses the x-axis.
Set y = 0 and solve for x:
0 = -3sin(-x+π/2)
Since sine of an angle equals zero when the angle is 0 or π, we can set -x+π/2 equal to these values and solve for x:
-x+π/2 = 0 or -x+π/2 = π
Solving the first equation:
x = π/2
Solving the second equation:
x = π/2 + π
x = 3π/2
So, the quarter points of y = -3sin(-x+π/2) are the ordered pairs (π/2, 0) and (3π/2, 0).
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3) Find the quarter points of y = 3cos(4x+π)+1:
Once again, we want to find the x-values where the function crosses the x-axis.
Set y = 0 and solve for x:
0 = 3cos(4x+π)+1
Now, subtract 1 from both sides of the equation:
-1 = 3cos(4x+π)
Divide both sides by 3:
-1/3 = cos(4x+π)
Since cosine of an angle equals -1/3 when the angle is arccos=(-1/3), we can set 4x+π equal to arccos=(-1/3) and solve for x:
4x+π = arccos(-1/3)
Solving for x, we have:
x = (arccos(-1/3) - π) / 4
There is no exact value for arccos(-1/3) using only π, so we'll leave the answer as (arccos(-1/3) - π) / 4.
Therefore, the quarter points of y = 3cos(4x+π)+1 are the ordered pairs [(arccos(-1/3) - π) / 4, 0].
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4) Find the quarter points of y = 2-3cos(4x-π):
Again, we want to find the x-values where the function crosses the x-axis.
Set y = 0 and solve for x:
0 = 2-3cos(4x-π)
Now, subtract 2 from both sides of the equation:
-2 = -3cos(4x-π)
Divide both sides by -3:
2/3 = cos(4x-π)
Since cosine of an angle equals 2/3 when the angle is arccos=(2/3), we can set 4x-π equal to arccos=(2/3) and solve for x:
4x-π = arccos(2/3)
Solving for x, we have:
x = (arccos(2/3) + π) / 4
There is no exact value for arccos(2/3) using only π, so we'll leave the answer as (arccos(2/3) + π) / 4.
Therefore, the quarter points of y = 2-3cos(4x-π) are the ordered pairs [(arccos(2/3) + π) / 4, 0].
I hope these step-by-step explanations help you understand how to solve these types of problems. Let me know if you have any further questions!