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C3 trig Question

Hi, i have another C3 question. Its on another topic so i started another thread (If that is that ok?)
https://7c95ed1f832973eac0c6e1a52b08181fd38e50f6.googledrive.com/host/0B1ZiqBksUHNYcWYycEk2VmM2b00/January%202013%20QP%20-%20C3%20OCR.pdf
Im an extremely lost on question 9(iii).
I looked at the market scheme and it said to "Attempt to relate k to at least 2 6sin 30".
I Cant see how this would help?
Thanks in advance. :smile:

(edited 7 years ago)
Reply 1
Avatar for Zacken
Zacken
22
Original post by SamuelN98
Hi, i got another C3 question. Its on another topic so i started another thread (Is that ok?)


I'm glad that you started another thread. :smile:

https://7c95ed1f832973eac0c6e1a52b08181fd38e50f6.googledrive.com/host/0B1ZiqBksUHNYcWYycEk2VmM2b00/January%202013%20QP%20-%20C3%20OCR.pdf
Im an extremely lost on question 9(iii).
I looked at the market scheme and it said to "Attempt to relate k to at least 2 6sin 30".
Cant see how this would help?


Can you see (i) of the question? Multiply through by 6 and compare the RHS. (the angles have to be divided by 1/3 as well)

We get k=6sin⁡2θ3k = 6\sin^2 \frac{\theta}{3}.

You know that since θ\theta is bounded between -90 and +90 then we have θ3\frac{\theta}{3} being bounded by -30 and +30.

So sin⁡θ3\sin \frac{\theta}{3} is bounded between -1/2 and +1/2.

So sin⁡2θ3\sin^2 \frac{\theta}{3} is bounded between 0 and 1/4.

So k=6sin⁡2θ3k = 6\sin^2 \frac{\theta}{3} is bounded between...?
Reply 2
Avatar for SamuelN98
SamuelN98
OP
5
Original post by Zacken
I'm glad that you started another thread. :smile:



Can you see (i) of the question? Multiply through by 6 and compare the RHS. (the angles have to be divided by 1/3 as well)

We get k=6sin⁡2θ3k = 6\sin^2 \frac{\theta}{3}.

You know that since θ\theta is bounded between -90 and +90 then we have θ3\frac{\theta}{3} being bounded by -30 and +30.

So sin⁡θ3\sin \frac{\theta}{3} is bounded between -1/2 and +1/2.

So sin⁡2θ3\sin^2 \frac{\theta}{3} is bounded between 0 and 1/4.

So k=6sin⁡2θ3k = 6\sin^2 \frac{\theta}{3} is bounded between...?


So, that gives the upper bound of 3/2 , how do you get the lower bound?
Reply 3
Avatar for Zacken
Zacken
22
Original post by SamuelN98
So, that gives the upper bound of 3/2 , how do you get the lower bound?


0.

0 * 6 is still 0.

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