Advertisements
Advertisements
प्रश्न
Evaluate: `cos (sin^-1 (4/5) + sin^-1 (12/13))`
Advertisements
उत्तर
`sqrt(5^2 - 4^2)` = 3
cos A = `"Adj"/"Hyp" = 3/5`
Let `sin^-1 (4/5)` = A
sin A = `4/5`
∴ cos A = `3/5`
`sqrt(169 - 144) = sqrt 25` = 5
cos B = `"Adj"/"Hyp" = 5/13`
Let `sin^-1 (12/13)` = B
`12/13` = sin B
sin B = `12/13`
∴ cos B = `5/13`
Now `cos (sin^-1 (4/5) + sin^-1 (12/13))` = cos (A + B)
= cos A cos B – sin A sin B
`= 3/5 xx 5/13 - 4/5 xx 12/13`
`= 15/65 - 48/65`
`= - 33/65`
APPEARS IN
संबंधित प्रश्न
In ΔABC, if a = 18, b = 24, c = 30 then find the values of cos `A/2`
In ΔABC, if a = 18, b = 24, c = 30 then find the values of tan `A/2`
Find the principal value of the following:
`sin^-1 (- 1/2)`
Solve: tan-1 (x + 1) + tan-1 (x – 1) = `tan^-1 (4/7)`
Show that `sin^-1 (- 3/5) - sin^-1 (- 8/17) = cos^-1 (84/85)`
Find the principal value of `cos^-1 sqrt(3)/2`
`sin^2(sin^-1 1/2) + tan^2 (sec^-1 2) + cot^2(cosec^-1 4)` = ______.
`sin[π/3 - sin^-1 (-1/2)]` is equal to:
`2"tan"^-1 ("cos x") = "tan"^-1 (2 "cosec x")`
If f'(x) = x–1, then find f(x)
