Trig question (cosine rule)

[Rogers1]

Hi. Just wondering if anyone could help me with this question? you'll find the question attached to this thread. Your help is very much appreciated :-D

I have so far found that the answer to (a) is somewhat like this:

c^2 = a^2 + b^2 -2b^2abcosC

let a = b

so: c^2 = 2b^2-2b^2b^2cosC

let c = ka
so: (ka)^2 = 2b^2 - 2b^2 cosC
k^2a^2 = 2b^2- 2b^2cosC
k^2a^2 - 2b^2= - 2b^2cosC

Let a = b

k^2b^2-2b^2 = -2b^2cosC
k^2b^2-2b^2/-2b^2 = cosC
b^2 (k^2-2)/ -2b^2 =cosC (cancel out b^2)
k^2-2/-2 (cancel negatives) so it will look like this : - (2-k^2)/-2

will look like this : 2-k^2/2

 

[Deleted member 4993]

Hi. Just wondering if anyone could help me with this question? you'll find the question attached to this thread. Your help is very much appreciated :-D

I have so far found that the answer to (a) is somewhat like this:

c^2 = a^2 + b^2 -2b^2abcosC

let a = b

so: c^2 = 2b^2-2b^2b^2cosC

let c = ka
so: (ka)^2 = 2b^2 - 2b^2 cosC
k^2a^2 = 2b^2- 2b^2cosC
k^2a^2 - 2b^2= - 2b^2cosC

Let a = b

k^2b^2-2b^2 = -2b^2cosC
k^2b^2-2b^2/-2b^2 = cosC
b^2 (k^2-2)/ -2b^2 =cosC (cancel out b^2)
k^2-2/-2 (cancel negatives) so it will look like this : - (2-k^2)/-2

will look like this : 2-k^2/2

For (b) and (c) - hint:

-1 ≤ cos(Θ) ≤ 1

then


-1 ≤ \(\displaystyle \dfrac{2-k^2}{2}\) ≤ 1