Problems and Quizzes related to Inverse Trigonometric Ratios
1. Problem : Find the principal value cot-1(-13)
2. Problem : Evaluate each of the following:
(a) cos (Arcsin 3/5),
(b) sin [Arccos (-2/3)],
(c) tan [Arcsin (-3/4)]
3. Problem : Show that
(i) sin-1 ( 2x1-x2) = 2 sin-1x, -12≤x≤ 12
(ii) sin-1 (2x1-x2) = 2 cos-1x, 12≤x≤1
4. Problem : Find the domain of sin-12x+π/6.
5. Problem : Express tan-1( cosx1-sinx ), -π2<x<π2 in the simplest form.
6. Problem : Find the value of sin-1 (sin 3π/5)
7. Problem : Show that sin-112/13 + cos-14/5 + tan-163/16 = π
8. Problem : Evaluate cos (Arctan 15/8 - Arcsin 7/25).
9. Problem : Solve tan-12x + tan-13x = π/4
10. Problem : Prove that 2 tan-113 + tan-117 = π/4
11. Problem : Solve the equation
tan-1x+1x-1 + tan-1x-1x = tan-1(-7)Solution : Here we have
a + b = cwhere
a = tan-1x+1x-1 => tana = x+1x-1
b = tan-1x-1x => tanb = x-1xand
c = tan-1(-7) => tanc = -7Since tan(a+b) = tanc
∴ tana+tanb1-tana.tanb = tanc
i. e. x+1x-1+x-1x1-x+1x-1.x-1x = -7
i.e. 2x2-x+11-x = -7
so that x=2
This value makes the left-hand side of the given equation positive, so that there is no value of x strictly satisfying the given equation.
The value x = 2 is a solution of the equation
tan-1x+1x-1 + tan-1x-1x = π + tan-1(-7)
12. Problem : Solve Arccos 2x - Arccos x = π/3 .
13. Problem : Graph
i) y = cos-1x + 1 ii) y = sin-1(x-2)
14. Problem : Show that one can use a composition of trigonometry buttons such as, sin, cos, tan, sin−1, cos−1, and tan−1, to replace the broken reciprocal button on a
15. Problem : A camera is placed on a deck of a pool. A diver is 18 feet above the camera lens. The extended length of the diver is 8 feet.
16. Problem : If sin-1 x + (sin-1 y + sin-1 z) = π/2, Find out x2 + y2 + z2 + 2xyz.
17. Problem : If sin-1(x- x22 + x34 ....) cos-1(x2 - x42 + x64 - ....) = π/2
18. Problem : Solve the equation:
tan-1 2x + tan-1 3x = nπ + π/4
19. Problem : Which angle is greater?
A = 2tan-1(22 - 1) and B = 3sin-1(13) + sin-1(35)Solution : We observe 22 - 1≈ 2(1.4) – 1 = 2.8 – 1 = 1.8
So 22 – 1 > 3 => 2tan-1(22 – 1) >2tan-13 = 2π3
B = 3sin-1(13) + sin-1(35)
= sin-1[3. 13 - 4(13)3] + sin-1(35)
= sin-1(2327) + sin-1(35)
< sin-1(32) + sin-1(32) = 2π3 [∵2327 < 32 and 35 < 3 2]So A > B.
20. Problem : Suppose a calculator is broken and the only keys that still work are the sin, cos, tan, sin−1, cos−1, and tan−1 buttons. The display initially shows 0. Given any positive rational number q, show that we can get q to appear on the display panel of the calculator by pressing some finite sequence of buttons. Assume that the calculator does real-number calculations with infinite precision, and that all functions are in terms of radians.
Solution: Because cos−1 sin θ = π2−θ and tan(π2−θ)= 1tanθ for 0 < θ < π2 , we have for any x > 0,
tan cos−1 sin tan−1 x = tan(π2− tan−1x)= 1x.............................. (*)Also, for x ≥ 0,
cos tan−1√x = 1x+1 ,so by (*),
tan cos−1 sin tan−1 cos tan−1√x =x+1............................. (**)By induction on the denominator of r, we now prove that √r, for every non-negative rational number r, can be obtained by using the operations
√x → x+1 and x → 1x.If the denominator is 1, we can obtain √r, for every nonnegative integer r, by repeated application of √x → x+1. Now assume that we can get√r for all rational numbers r with denominator up to n. In particular, we can get any of
n+11, n+12, ... n+1n,so we can also get
1n+1, 2n+1, ... nn+1,and
√r, for any positive r of exact denominator n + 1, can be obtained by repeatedly applying √x → x+1.
Thus for any positive rational number r, we can obtain √r. In particular, we
can obtain q2 = q.
Complete Tutorial (MCQ Quizzes after this):
Companion MCQ Quiz #2 for Inverse Trigonometric Ratios (Challenging Problems) - test how much you know about the topic. Your score will be e-mailed to you at the address you provide.