MEDALS! Question 8 without using angle sum identities. This is a screening test for a summer course for linear algebra in Hong Kong University. I was lucky as I thought that they would ask questions on trigonometry so I memorised the angle sum identities in the last minute. I think I got this right and I am curious on how to do question 8 without the identities.
Question 8:
@BSwan
why dnt you type a question at once xD :'(
question 1 \(\sum_{n=1}^{n} (n+1) 2^{n-1}=n2^n\) by induction :- \(\color{red}{\text {STEP 1:-}}\) for n=1 2=2 its correct \(\color{red}{\text {STEP 2 :-}}\) assume its true for k \(\sum_{k=1}^{k} (k+1) 2^{k-1}=k2^k\) \(\color{red}{\text {STEP 3 :-}}\) show its true for k+1 :P can u do it ?
the question is to evaluate sin(p+q) why would you wanna go long way though
@BSwan he wants question 8 :P
sorry for the wasted efforts xD
O.O
i hate u @aajugdar
whaaa he mentioned he wants question 8 xD you didnt read
I finally got question 6 right after three trials. *sigh* Wasted quite a lot of time... :(
haha u gave right answer to a wrong question @BSwan :P
I mean in the test.
:D
its my bad i dint read lol
so stil wanna 8 ?
@BSwan he wants to solve it without using trigonometry
andd i hv no clue why o_o
Yep. I want to know how people can solve this without the identity.
I was lucky as I memorised the identity in the last minute (literally five minutes before the test). I just want to know how others can solve it.
wew ! |dw:1401549520202:dw|
so the bottom triangle is a 3,4,5 triangle
I think the tricky bit is how to draw a triangle with p+q as one of the angles.
is m parallel to f ? |dw:1401549742942:dw|
But not everyone in the test knows the identity so there must be a way solving it without using the identity. (ah... I know it is faulty logic but I just want the answer lolololol)
yes @BSwan there is no harm in constructing a parallel line to f, its perfectly valid
:D |dw:1401549900070:dw|
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