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Question
Show that:
`cos^(-1)(4/5)+cos^(-1)(12/13)=cos^(-1)(33/65)`
Solution
Let a = `"cos"^-1 (4/5)` and b = `"cos"^-1 (12/13)`
Let a = `"cos"^-1 (4/5)`
cos a = `4/5`
We know that
sin2a = 1 - cos2a
sin a = `sqrt (1-"cos"^2 "a")`
`= sqrt (1 - (4/5)^2) = sqrt (1 - 16/25)`
`= sqrt ((25-16)/25) = sqrt (9/25) = 3/5`
Let b = `"cos"^-1 (12/13)`
cos b = `12/13`
W know that
sin2b = 1 - cos2b
sin b = `sqrt (1 - "cos"^2 "b")`
`= sqrt (1 - (12/13)^2) = sqrt (1 - 144/169)`
`= sqrt ((169-144)/169) = sqrt (25/169) = 5/13`
We know that
cos (a+b) = cos a cos b - sin a sin b
Putting values
cos a = `4/5` , sin a = `3/5`
& cos b = `12/13` , sin b = `5/13`
cos (a+b) = `4/5 xx 12/13 xx 3/5 xx 5/13`
`= 48/65 - 3/13`
`= (48 - 15)/65`
`= 33/65`
∴ cos (a+b) = `33/65`
a + b = cos-1 `(33/65)`
`"cos"^-1 4/5 + "cos"^-1 (12/15) = "cos"^-1 (33/65)`
Hence LH.S = R.H.S
Hence proved.
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