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#1
If I have the equation y = -10 cos(-8[x-6])
And I have to work out amplitude, period and phase shift

For amplitude 10
Period I'm a bit unsure, I did -8 x 1 which is -8 (as there is a 1 in front of the x) and so it's - 2Pi/8 which is - Pi/4 is this correct?
0
2 weeks ago
#2
(Original post by Helloaaa)
For amplitude 10
Period I'm a bit unsure, I did -8 x 1 which is -8 (as there is a 1 in front of the x) and so it's - 2Pi/8 which is - Pi/4 is this correct?
Yes, it's correct except a period is a positive quantity (time between repeats) so a negative answer doesn't make sense, so pi/4.
Last edited by MarkFromWales; 2 weeks ago
0
#3
(Original post by MarkFromWales)
Yes, it's correct except a period is a positive quantity (time between repeats) so a negative answer doesn't make sense, so pi/4.
Oh yes!! Thanks for pointing that out. How about for the phase shift we wouldn't have to multiply the brackets out for that right? So it would be x-6=0 so phase shift is 6 is that correct?
0
2 weeks ago
#4
(Original post by Helloaaa)
Oh yes!! Thanks for pointing that out. How about for the phase shift we wouldn't have to multiply the brackets out for that right? So it would be x-6=0 so phase shift is 6 is that correct?
Yes, that's correct .
0
#5
(Original post by MarkFromWales)
Yes, that's correct .
If I have Y1 (x) = 3 sin (x) and y2 (x) = -9 sin (x)

Would the sum be 3 sin(x) + -9 sin(x) so the answer is -6?
0
2 weeks ago
#6
(Original post by Helloaaa)
If I have Y1 (x) = 3 sin (x) and y2 (x) = -9 sin (x)

Would the sum be 3 sin(x) + -9 sin(x) so the answer is -6?
The sum of 3 sin(x) and -9 sin(x) is -6 sin(x).
(the amplitude of that is 6)
0
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