SOLVING PROBLEMS WITH INVERSE TRIG FUNCTIONS

Problem 1 :

The value of sin^-1 (cos x) = 0, 0 ⩽ x ⩽ π is

1) π - x     2)  x - (π/2)      3)  (π/2) - x       4)  x - π

Solution :

sin^-1 (cos x) = 0

cos x = sin 0

cos x = 0

cos x = sin ((π/2) - x)

So, the answer is (π/2) - x.

Problem 2 :

If sin^-1 x + sin^-1y = 2π/3, then cos^-1 x + cos^-1y is equal to 

1) 2π/3     2)  π/3      3)  π/6       4)  π

Solution :

sin^-1 x + sin^-1y = 2π/3

π/2 - cos^-1 x + π/2 - cos^-1 y = 2π/3

- cos^-1 x - cos^-1 y = 2π/3 - π/2 - π/2

-(cos^-1 x + cos^-1 y) = 2π/3 - π

cos^-1 x + cos^-1 y = π - (2π/3)

cos^-1 x + cos^-1 y = (3π - 2π)/3

cos^-1 x + cos^-1 y = π/3

So, the value of cos^-1 x + cos^-1 y is π/3.

Problem 3 :

Solution :

We know the formula,

sin^-1x + cos^-1y = π/2

cosec^-1x = 1/sin^-1x

sec^-1x = 1/cos^-1x

Problem 4 :

If sin^-1 x = 2 sin^-1α has a solution, then

1) |α| ≤ 1/√2     2)  |α| ≤ 1/√2    3)  |α| < 1/√2      4)  |a| > 1/√2

Solution :