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Trignometry

Blazing goIITian

Joined:	31 Jan 2007
Post:	515

18 Jul 2007 17:21:12 IST

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1103

Inverse trignometry URGENT!!!!!!!!!!!!

None

If sin^-1 x/a + cos^-1 y/b = , then show that x² / a²+ y²/ b² - 2xy/ ab sin . = cos² .

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rashmi varma

Hot goIITian

Joined: 8 Mar 2007

Posts: 184

18 Jul 2007 18:06:39 IST

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hi varsha.

change the cos to sin and apply formula of sin-(c)+sin-(d)...(- mean inverse)

u get

sin-(x/a)+sin-(sqrt(b2-y2)/b)

solve it u"ll get

xy/ab+^{[ ]}

b²-y² *^{[ ]}

a²-x²/ab=sin(alpha)

take ab to right side

xy+^{[ ]}

b²-y² *^{[ ]}

a²-x²=sin(alpha)ab......(1)

square both sides and substitute value of the term

^{([ ]}

b2-y2.....)*2xy from 1.

solve it.

u"ll get da required result.

Hope its clear.Actually I'm a bit lazy to write all the steps.But still if u can't get I"ll do that.

Buhbyee

Titun

Forum Expert

Joined: 23 Dec 2006

Posts: 374

18 Jul 2007 18:15:56 IST

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Let sin ^-1(x/a) = m

x / a = sin m -

/ 2 < = m < = pie / 2
Let cos ^-1(y/b) = z

y / b = cos z 0 < = z < = pie

sin ^-1(x/a) + cos ^-1(y/b) =

m + z =

Therefore, cos ²

= cos ² (m+z) = (cos m.cos z - sin m. sin z )²
              = cos²m . cos ²z + sin ² m.sin ²z - 2cos m.cos z.sin m.sin z
              = cos² z ( 1- sin²m) + sin²m (1-cos²z) - 2cos m.cos z.sin m.sin z
              = cos²z + sin²m - 2sin m . cos z. (sinz.cosm + cosz.sin m)
              = cos²z + sin²m - 2sin m . cos z. sin (m+z)
              = y²/b² + x²/a² - 2xy / ab . sin

Hence, proved.

varsha valli g.

Blazing goIITian

Joined: 31 Jan 2007

Posts: 515

19 Jul 2007 16:51:01 IST

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thank u very much titun & rshm15varma.

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