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Question
Prove the following:
`sin^-1(3/5) + cos^-1(12/13) = sin^-1(56/65)`
Solution
Let `sin^-1(3/5) = x, cos^-1(12/13) = y and sin^-1(56/65)` = z.
Then `sin x = (3)/(5), "where" 0 < x < pi/(2)`
cos y = `(12)/(13), "where" 0 < y < pi/(2)`
and sin z = `(56)/(65), "where" 0 < z < pi/(2)`
∴ cos x > 0, sin y > 0
Finding sin x, cos x
Now, cos x = `sqrt(1 - sin^2 x)`
= `sqrt(1 - ((3)/(5))^2 `
= `sqrt(1 - (9)/(25)` = `sqrt(16/25) = (4)/(5)`
Finding sin y, cos y
sin y = `sqrt(1 - cos^2y)`
= `sqrt(1 - ((12)/(13))^2`
= `sqrt(1 - (144)/(169)` = `sqrt(25/169) = (5)/(13)`
We know that
`sin(x + y) = sin x cos y + cos x sin y`
= `(3/5) "x" (12/13) + (4/5) "x" (5/13)`
= `(36)/(65) + (20)/(65) = (56)/(65)`
`∴ sin(x + y) = (56)/(65)`
`∴ x + y = sin^-1 (56)/(65)`
Hence, `sin^-1(3/5) + cos^-1(12/13) = sin^-1(56/65)`.
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