मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Show that 9π8-94sin-1(13)=94sin-1(223).

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प्रश्न

Show that `(9pi)/8 - 9/4 sin^-1 (1/3) = 9/4 sin^-1 ((2sqrt2)/3)`.

बेरीज
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उत्तर

We have to show that

`(9pi)/8 - 9/4 sin^-1 (1/3) = 9/4 sin^-1 ((2sqrt2)/3)`

i.e. to show that,

`9/4 sin^-1 (1/3) + 9/4 sin^-1 ((2sqrt2)/3) = (9pi)/8`

Let `sin^-1 (1/3)` = x

∴ sin x = `1/3, "where"  0 < x < pi/3`

∴ cos x > 0

Now, `cos "x" = sqrt(1 - sin^2"x") = sqrt(1 - 1/9) = sqrt(8/9) = ((2sqrt2)/3)`

∴ x = `cos^-1 ((2sqrt2)/3)`

∴ `sin^-1 (1/3) = cos^-1 ((2sqrt2)/3)`    ....(1)

∴ LHS = `9/4 sin^-1 (1/3) + 9/4 sin^-1 ((2sqrt2)/3)`

= `9/4 [sin^-1 (1/3) + sin^-1 ((2sqrt2)/3)]`

`= 9/4 [cos^-1 ((2sqrt2)/3) + sin^-1 ((2sqrt2)/3)]` ...[By (1)]

`= 9/4 (pi/2)  ....[because sin^-1"x" + cos^-1"x" = pi/2]`

`= (9pi)/8`

= RHS.

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पाठ 3: Trigonometric Functions - Miscellaneous exercise 3 [पृष्ठ ११०]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 3 Trigonometric Functions
Miscellaneous exercise 3 | Q 22 | पृष्ठ ११०

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