Mathematics Skill Development - Module 7
Mathematics Skill Development - Module 7
Inverse Trigonometric Functions
The following questions will evaluate the student’s understanding of inverse trigonometric functions.
1. Find the exact value of
√ !
3
−
.
2
−1
sin
Solution:
Since sin(π/3) =
hence
√
√
3/2 and sin is an odd function, sin(−x) = − sin(x), we get sin(−π/3) = − 3/2 and
√ !
3
π
−
=− .
2
3
−1
sin
2. Evaluate
cos−1
π
3
.
Solution:
We know that −1 ≤ cos(x) ≤ 1 yet
π
3
> 1. Therefore, there cannot exist a value of x such that
cos(x) =
π
> 1,
3
and cos−1 (π/3) is not defined.
3. Find the exact value of
3
cos sin−1
.
5
Solution:
Let θ = sin−1 (3/5) so that sin(θ) = 3/5. Using trigonometric ratios, we construct a right-angled triangle
with these proportions:
5
3
θ
4
We used the Pythagoras theorem to compute the length of the side adjacent to the angle θ. Using the
trigonometric ratio for cos(θ), we have
3
4
cos sin−1
= cos(θ) = .
5
5
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Mathematics Skill Development - Module 7
4. Evaluate
sin(2 tan−1
1
.
2
Solution:
Let θ = tan−1 (1/2) so that tan(θ) = 1/2. Using the trigonometric ratio, we construct a right-angled
triangle with these proportions:
√
5
1
θ
2
where we used the Pythagoras theorem to compute the length of the hypotenuse to be
sin θ = √15 and cos θ = √25 . Using the double angle formula, we compute
sin(2 tan−1
√
5. Therefore
1
= sin(2θ)
2
= 2 sin(θ) cos(θ)
1
2
=2· √ · √
5
5
4
= .
5
5. Evaluate and simplify the following expression
3
3
−1
−1
+ cos
.
cos sin
5
5
Solution:
Let α = sin−1 (3/5) and β = cos−1 (3/5) so that sin(α) = 3/5 and cos(β) = 3/5. Using the corresponding
trigonometric ratios as in Questions 3 and 4, we construct the following right angled triangles
5
5
3
β
3
α
4
4
Where we used the Pythagoras theorem to compute the missing side in each triangle. Using this diagram
and the addition formula for cos(α + β), we compute
cos(sin−1 (3/5) + sin−1 (3/5)) = cos α cos β − sin α sin β
4 3 3 4
= · − ·
5 5 5 5
= 0.
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Mathematics Skill Development - Module 7
It is also easy to see that the triangles above are exactly the same with the angles α and β located in
opposite corners. Since this is a right-angled triangle and angles add up to π, we must have α + β = π2 and
cos(α + β) = 0.
3