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Discussion Response Post to: Inverse trignometry URGENT!!!!!!!!!!!!

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Titun (1467)

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total posts: 374
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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.

You never know what is enough till you know what is more than enough. Titun

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