# C2 urgent trig question help

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#1

I need help with question 11. For part (a) I managed to answer it by trial and error. I need someone to explain how to do these questions. One thing I do know is the that the values of cos and sin are between -1 and 1.
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5 years ago
#2
(Original post by Butterflyshy)

I need help with question 11. For part (a) I managed to answer it by trial and error. I need someone to explain how to do these questions. One thing I do know is the that the values of cos and sin are between -1 and 1.
The minimum value of both and is -1, similarly the maximum value of both is 1. How can we use that information to determine a minimum and maximum value of y, and hence the values for x for which this value of y occurs?
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#3
(Original post by _gcx)
The minimum value of both and is -1, similarly the maximum value of both is 1. How can we use that information to determine a minimum and maximum value of y, and hence the values for x for which this value of y occurs?
Sorry to bother but can you help me go through part b.
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5 years ago
#4
(Original post by Butterflyshy)
Sorry to bother but can you help me go through part b.
The minimum value of is -1. How can we use this to get a minimum value of y? (try substituting)
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#5
(Original post by _gcx)
The minimum value of is -1. How can we use this to get a minimum value of y? (try substituting)
how did you get sin(3x-45)
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5 years ago
#6
(Original post by Butterflyshy)
how did you get sin(3x-45)
He confused (c) and (b). Anyway for (b), you have something of the form where . Now you want to make y as big as possible. Since y is of the form 5 - 4a, you want to then make 4a as small as possible since then you're subtracting the smallest number possible from 5, hence giving you the biggest y possible.

Now to make 4a as small as possible, you simply need to see how small a can get, then multiply that value by 4 to see how small 4a can get.

Since a = sin(x+30), which at it's very smallest is -1. Then 4a at it's very smallest is -4.

So y = 5 - 4a is at it's biggest when y = 5-(-4) = 9. That's the maximum.

Similarly to make y as small as possible, we want to make a as big as possible so that you're subtracting the biggest number possible from 5 to get the smallest y possible. Blah blah blah...

can you see the general techniques here, all it relies on is the fact that sin and cos are bounded between -1 and 1 and then using common sense to find the maximum and minimum of functions involving sines and cosines.

Let me know if this needs more clearing up...
2
#7
(Original post by Zacken)
He confused (c) and (b). Anyway for (b), you have something of the form where . Now you want to make y as big as possible. Since y is of the form 5 - 4a, you want to then make 4a as small as possible since then you're subtracting the smallest number possible from 5, hence giving you the biggest y possible.

Now to make 4a as small as possible, you simply need to see how small a can get, then multiply that value by 4 to see how small 4a can get.

Since a = sin(x+30), which at it's very smallest is -1. Then 4a at it's very smallest is -4.

So y = 5 - 4a is at it's biggest when y = 5-(-4) = 9. That's the maximum.

Similarly to make y as small as possible, we want to make a as big as possible so that you're subtracting the biggest number possible from 5 to get the smallest y possible. Blah blah blah...

can you see the general techniques here, all it relies on is the fact that sin and cos are bounded between -1 and 1 and then using common sense to find the maximum and minimum of functions involving sines and cosines.

Let me know if this needs more clearing up...

Thanks so much
0
5 years ago
#8
(Original post by Butterflyshy)
Thanks so much
No problem.
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#9
(Original post by Zacken)
No problem.
I now understand where the values of y occur but how would find the values of x. For example the max and min for part a are 2 and 0 and they occur at 90 degrees and 180 degrees?
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5 years ago
#10
(Original post by Butterflyshy)
I now understand where the values of y occur but how would find the values of x. For example the max and min for part a are 2 and 0 and they occur at 90 degrees and 180 degrees?
Well to find the max of y, you had to set sin (x+30) = 1. Now you solve this equation for x.

Like wise for min and sin(x+30) = -1
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#11
(Original post by Zacken)
Well to find the max of y, you had to set sin (x+30) = 1. Now you solve this equation for x.

Like wise for min and sin(x+30) = -1
Thanks I'm getting the right answer.
0
5 years ago
#12
(Original post by Butterflyshy)
Thanks I'm getting the right answer.
Great.
0
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