4 - Inverse Trigonometry Questions Answers
sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6) = ?
sin‾¹(sin2) + sin‾¹(sin4) + sin‾¹(sin6)
= (π-2) + (π-4) + (6-2π)
= 0
If cos-1x - cos-1(y/2) = α , then 4x2-4xycosα + y2 is equal to ?
according to the given condition
cos-1x = cos-1(y/2) + a
so x = cos ( cos-1y/2 + a)
so x = y/2 cosa - sin cos-1y/2 + sina
so x = y/2 cosa - [1-(y2/4)]1/2 + sina
so x - sina - y/2 cosa = -[1-(y2/4)]1/2
now square both side and solve
If sin-1a + sin-1b+ sin-1c = π , then the value of { a√(1-a2) + b√(1-b2) + c√(1-c2)}
let sin-1a = A , sin-1b = B and sin-1c= C
so { a√(1-a2) + b√(1-b2) + c√(1-c2)} = sinAcosA + sinBcosB + sinCcosC = 1/2 (sin2A+sin2B+sin2C)
= 1/2 (4sinAsinBsinC)
= 2abc
If [cot‾1x] + [cos‾1x] = 0 , then complete set of value of x is ( [ * ] is GIF)?
These are the graphs for cot-1x and cos-1x
violet for cos and green for cot
from graph it is clear that cos part with integer function will be 0 for cos1<x≤1
and cot part with integer function will be 0 for cot1<x<∞
so answer will be cot1<x≤1