Question

Fool's problem of the day,

Today a very easy trigonometry problem,

\( (1) \) If \( \cos ^{-1} x - \cos ^{-1} \frac y 2 = \alpha \), then find the value of \( 4x^2-4xy\cos \alpha +y^2\)

PS:The answer is (probably) Google-able, so post a solution instead ;)

Answer 1

When you all get done here please come help me with my question. Nobody seems to be helping anymore.

Answer 2

\[\cos (\cos^{-1} x-\cos^{-1} y \div2)=\cos \alpha\]
\[x \times y \div 2 + {(\pi-2x)\times(\pi-y)}\div4 =\cos \alpha\]

Answer 3

its easy
cos(inv) [x*y/2 + sqrt((1-x2)(1-y2/4))]=a
sqrt((1-x2)(1-y2/4))]=cos a - x*y/2 ,now square both sides
(1-x2)(1-y2/4)=cos^2 a + ((xy)^2)/4 - xycosa
1-y2/4 -x2 +(xy)^2/4=cos^2 a + ((xy)^2)/4 - xycosa
1-y2/4 -x2=cos^2 a - xycosa
4-y2-4x2=4cos ^2 a - 4xycosa
4x2-4xycosa+y2=4(1-cos^2 a)=4sin^2 a
Q.E.D.

Answer 4

is it okay @experimentX @FoolForMath

Answer 5

hahah ... lol you seriously overrate me.

Answer 6

"overrate"-what's that supposed to mean???

Answer 7

like i would know the answer ... how did you get the first step??

Answer 8

i tried many ways on paper but this method stood out

Answer 9

Congratz @Sarkar You did it! :D

Answer 10

I didn't understand the first step??

Answer 11

I am thinking to make a Hall of Fame for the problem of the day, What say you guys?

Answer 12

yeah yeah. sure!

Answer 13

I understand ... now.

Answer 14

@experimentX - i just used the formula of
cos(inv) x - cos(inv) y= cos(inv)[xy-sqrt(1-x2)(1-y2)]

Answer 15

cos(inv) x - cos(inv) y= cos(inv)[xy+sqrt(1-x2)(1-y2)]

Answer 16

Yes, that's Inverse circular function 101!

Answer 17

wow ... there was a formula like that???
Congratz @Sarkar
Taking Cos(...) = cos(a) yields the same.

Answer 18

I took cos ... got your second step ... and stuck .. lol

Answer 19

Yes @experimentX It's easy to prove too, Consider \( cos^{-1} x = A\) and \(\cos^{-1} y = B\) and then use \( \cos (A+B) \)

Answer 20

Stuck why???@X

Answer 21

hahaha ... I'll remember that.
and also persistence is the key to success.

Answer 22

Inverse circular function doesn't have much significance higher mathematics (I don't know why!) so it's somewhat least practiced.

Answer 23

Well, I never learned that on school ... ;D

Answer 24

i do agree @FFM but its in our course so have to brush it constantly...

Answer 25

It's standard high school material here.

Answer 26

@Sarkar JEE aspirant?

Answer 27

yes@FFM

Answer 28

and you???

Answer 29

2013?

Answer 30

yup...

Answer 31

I failed long back :P

Answer 32

Good luck! :)

Answer 33

Good Luck!!

Answer 34

are you an Indian.???

Answer 35

@FFM

Answer 36

thanks for your wishes

Answer 37

Yes of-course :)

Answer 38

didnt know that until now,,lol

Answer 39

lol, I don't usually publicize my nationality unless asked to do so ;)

Answer 40

whatever,can you help me with a sum???

Answer 41

Foolformath can help you with ANY sum!! :)

Answer 42

Okay guys, hope you all enjoyed today's problem. See you all very soon.

@Sarkar Sorry, I have to leave now, you can you post it via "Ask the question..." feature. I will check it later :) Just don't forget to type @FoolForMath. Since, @FFM don't notify me.

Answer 43

ohh.. ffm is some other username. i see. lol

Answer 44

be back with Hall of Fame problem!!!

Answer 45

it goes something like this
\[\sin^{-1} [(a+bcosx)/(b+acosx)]\]...diffrentiate it
i know how to diff it but is there a possible sub we can use here????

Answer 46

will do @FoolForMath

Answer 47

@Sarkar if you put that into b/a - ((b^/a)-b)/(b+acosx) .... it might make simpler.

Answer 48

* b/a - ((b^2/a)-b)/(b+acosx)