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secondary 4 | A Maths
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Sonia
Sonia

secondary 4 chevron_right A Maths chevron_right Singapore

Hi! Good afternoon !
Can someone help me with all of these parts ? especially for the period and equation as i can’t tell whether it’s a sine or cosine graph ..
Thanks so much :DD

Date Posted: 4 years ago
Views: 226
Eric Nicholas K
Eric Nicholas K
4 years ago
Hi Sonia! I will look at this question later at night.

I give you a clue first: it’s a cosine graph since the graph cuts the y axis at its maximum value.
Arnold K H Tan
Arnold K H Tan
4 years ago
Does not matter whether sine or cosine. Period = angle for one complete wave = 6pi - (-2pi) = 8 pi
Amplitude = 2 (maximum displacement from x-axis in this case, or (highest point - lowest point) ÷ 2
Eric Nicholas K
Eric Nicholas K
4 years ago
But sine comes with a phase shift and its graph is not exactly covered in detail. I presume the question is to be asked as a sin kx or cos kx version rather than sin (kx + alpha).

Sonia, I will write this question down for you with explanations by 4 am.
J
J
4 years ago
For the equation ,

y = 2cos(x/4)
or
y = 2sin(π/2 - x/4)
or
y = 2sin(x/4 + π/2)



For the last part,


k + 8π = m

Or

2cos((k+8π)/4) = 2 cos(k/4 + 2π) = 2cos(m/4)
J
J
4 years ago
Btw, is that -1.5 changed on purpose? I would think it should be -√2 instead
Eric Nicholas K
Eric Nicholas K
4 years ago
Hard to say. That’s a possibility.

I will analyse that graph in detail later.
Eric Nicholas K
Eric Nicholas K
4 years ago
I presume y = -1.5 is meant to be correct, and that k is actually between -4pi and -3pi. Upon closer inspection of the picture, k is very slightly positioned to the left of -3pi.

I have done a cross-analysis in Desmos with the equation y = 2 cos (0.25pi x) and found that the point where y = -1.5 is when x = -3.0798.

So, in the actual graph y = 2 cos (0.25x), k should be approximately -3.0798pi.
J
J
4 years ago
No wonder. Desmos gave -√2 for x = -3π/ 5π. Guess one had to take a closer look at the question's graph. Without zooming in it looked like it was at 5π/-3π

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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
Good evening Sonia!!!

For this curve, at your level, you will only be required to sketch graphs of y = a sin bx + c and y = a cos bx + c, rather than y = a sin (bx + c) + d or y = a cos (bx + c) + d.

As such, for graphs starting at the centre point from the y-axis, it would be a sine curve. Specifically, we have a positive sine curve if the curve goes up first and a negative sine curve if the curve goes down first.

For graphs starting at the maximum point from the y-axis (heading downwards), it would be a positive cosine curve. For graphs starting at the minimum point from the y-axis (heading upwards), it would be a negative cosine curve.

Let me know if you need more explanation and I will do my best to explain them again!
J
J
4 years ago
Makes me wonder if 2sin(x/4 + ½π) would be marked correct by Cambridge