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Trignometry

Blazing goIITian

Joined:	16 Apr 2007
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27 Jul 2007 21:31:15 IST

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709

INVERSE TRIGNO-ans req soooooooooooooooooonnnnnnnnnnnnnnnnnnnn

None

1)cos^-1x₁+cos^-1x₂+......+cos^-1x_n=npi, then find the value of
x_1²+x₂²+x²₃+....+x_{n^{2

this is the way i solved it
cos-1x<=pi
so x=-1}}x_1²+x₂²+x²₃+....+xn2=1+1+1+...=n_{^{PLS confirm theANS
2)sin}^-1x1+}sin^-1x2+...+sin^-1x2n=npi

find x1+x2+x3...+x2n

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Comments (6)

Akansha

Blazing goIITian

Joined: 16 Apr 2007

Posts: 663

28 Jul 2007 00:03:18 IST

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pls reply !!

Titun

Forum Expert

Joined: 23 Dec 2006

Posts: 374

28 Jul 2007 14:30:41 IST

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1st problem.

0

cos ^{- 1} x

Therefore, 0

cos ^-1x ₁

cos ^-1x ₂

                         .
                         .
                         .
                         .
                         .
              0

cos ^-1x _n

Therefore,

cos ^-1x₁ + cos ^-1x₂ + ................ cos ^-1x_n

The equality holds only when

cos ^-1x₁ = cos ^-1x₂= .............................. = cos ^-1x_n =

i.e x₁ = x₂ = ..........................................= x_n= - 1

So,
x₁² + x₂² + ............................+ x_n² = 1+1+ .............n times = n

2nd problem.

-

/ 2

sin ^{- 1} x

/2

Therefore, -

sin ^-1x ₁

/2
-

sin ^-1x ₂

/2
                         .
                         .
                         .
                         .
                         .
             -

sin ^-1x _2n

/2

Therefore,

sin^-1x₁ + sin ^-1x₂ + ................ sin ^-1x_2n

2n .

/2 = n

The equality holds only when

sin ^-1x₁ = sin ^-1x₂= .............................. = sin ^-1x_2n =

/2
i.e x₁ = x₂ = ..........................................= x_2n= 1

So,
x₁ + x₂ + ............................+ x_2n = 1+1+ .............2n times = 2n

Cheers !!!!!

Akansha

Blazing goIITian

Joined: 16 Apr 2007

Posts: 663

28 Jul 2007 15:07:17 IST

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thanx fr the reply titun bt the ANS FR THE 2ND 1 IS WRONG
ANS :0 FR THE 2ND ONE

Jatin Sharma

Blazing goIITian

Joined: 20 Jun 2007

Posts: 323

28 Jul 2007 22:33:58 IST

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It is a simple question.

We are aware of the fact that the range of cos-1x is 0 to pi

The equation will hold only when each of the term individually is pi.

Hence x1 = x2 =...xn = -1

Hence when you sum them all, we get n

Jatin Sharma

Blazing goIITian

Joined: 20 Jun 2007

Posts: 323

28 Jul 2007 22:37:20 IST

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Similarly for the second question,

the range of sin^-1x is -pi/2 to pi/2

Hence they will all individually have to be 1.

Hence the answer is 2n.

sinjan jana

Blazing goIITian

Joined: 7 Jun 2007

Posts: 480

1 Aug 2007 18:56:54 IST

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here it is

Problem 1:

Assuming values in [0,\pi] for cos inverse function, the only way to add n positive numbers to get n\pi is to have each one of them as pi. So, the logic looks sound to me.

Problem 2 Hint:

Assume values in $[-\frac{\pi}{2},\frac{\pi}{2}]$ and try to proceed. Also, note that there are 2n terms here.

i am sure it will help u to do the sum independently.

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