Question

Hiii guys✌✌✌

$y=cos{}^{1}(cos10) also X=sin{}^{-1}( sin10) then y-x?$

Note:- answer is not zero

Need perfect answer

Answer 1

y = cos^-1(cos10) and x = sin^-1(sin10)

answer is not zero , it means here 10 is in radian not in degree.

so, 10 = 10π/π = 10π/(22/7) = 70π/22 = 35π/11 =(33π + 2π)/11 = 3π + 2π/11

we know, y = cos^-1(cosA) = A if 0 < A < π

see graph of cos^-1(cosx)
here it is clear that, graph between 3π to 4π
so, equation of line between 3π to 4π :
Y - π = (0 - π)/(4π - 3π)(X - 3π)
Y - π + (X - 3π) = 0
X + Y - 4π = 0
so, Y = 4π - X

so, y = cos^-1(cos10) = 4π - 10

again, for x = sin^-1(sin10) see graph, between 3π to 7π/2 [ because 3π + 2π/11 lies between them ]

here, equation of line lies between 3π to 7π/2
points are (5π/2, π/2) and (7π/2, -π/2) is
Y + π/2 = (-π/2 - π/2)/(7π/2 - 5π/2)(x - 7π/2)
Y + π/2 + (X - 7π/2) = 0
Y + X - 3π = 0
Y = 3π - X

hence, x = sin^-1(sin10) = 3π - 10

now, y - x = 4π - 10 - (3π - 10) = π

hence, answer should be π

Answer 2

Hey dear,
That's an interesting question. If 10 is in degrees answer will be 0. But I suppose you are asking in radians.

◆ Answer -
y - x = π

◆ Explaination-
We know, range of sininv(m) is [-π/2,π/2] where m E (-1,1).

As 10 doesn't lie in [-π/2,π/2] but 3π-10 does.
Hence,
x = sininv[sin(10)]
x = sininv[sin(3π-10)]
x = 3π-10


We know, range of cosinv(m) is [0,π] where m E (-1,1).

As 10 doesn't lie in [0,π] but 4π-10 does.
Hence,
y = cosinv[cos(10)]
y = cosinv[cos(4π-10)]
y = 4π-10


Therefore,
y - x = 4π-10 - (3π-10)
y - x = π

Hope this helps...