Inverse Trigonometric Functions — Challenge
Sine repeats forever, so its inverse has to be told which answer to give back.
What these problems cover
These questions use 12 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Repeating — Every fraction becomes a decimal that either stops or repeats forever.
- Function — A rule pairing each input with exactly one output.
- Range — The set of outputs that actually occur.
- Adjacent — The side next to that angle which is not the hypotenuse.
- Sine — sin θ = oppositehypotenuse.
- Cosine — cos θ = adjacenthypotenuse.
- Composition — Running one function on the output of another, written (f ° g)(x) = f(g(x)).
- Reference angle — The acute angle between the terminal side and the x-axis.
- Arcsine — The inverse of sine, restricted to outputs from -90° to 90°.
- Arccosine — The inverse of cosine, restricted to outputs from 0° to 180°.
- Arctangent — The inverse of tangent, restricted to outputs strictly between -90° and 90°.
- Quadrant — One of the four regions the axes divide the plane into.
Problem 1
Evaluate arccos(cos 300°).
- 60°
- 300°
- -60°
- 120°
- 240°
Worked answer
60°
cos 300° = 12, since the reference angle is 60° and cosine is positive in quadrant IV. Arccosine must return an angle between 0° and 180°. The angle in that range with cosine 12 is 60°. It cannot hand back 300°, however much you might want it to.
Problem 2
How many real solutions does arcsin x = arccos x have?
- 0
- 1
- 2
- Infinitely many
- Cannot be determined
Worked answer
1
If the two inverses give the same angle θ, then sinθ = cosθ = x. That happens at θ = 45°, where both are √22. So x = √22 is the only solution — the ranges overlap only on [0°, 90°], where sin rises and cos falls, crossing exactly once.
Problem 3
Evaluate sin(arccos35).
- 45
- 35
- 53
- -45
- 54
Worked answer
45
Let θ = arccos35, so cosθ = 35. By the Pythagorean identity sin2θ = 1 - 925 = 1625. Arccosine returns angles from 0° to 180°, where sine is never negative, so sinθ = 45.
Problem 4
Evaluate tan(arcsin513).
- 512
- 125
- 135
- 513
- -512
Worked answer
512
Let θ = arcsin513. Picture a right triangle with opposite 5 and hypotenuse 13. The third side is 12, and arcsine returns an angle in [-90°, 90°] where cosine is positive, so the adjacent side counts as +12. Then tanθ = 512.
Problem 5
For every x in [-1,1], what is arcsin x + arccos x?
- 90°
- 180°
- 0°
- It depends on x
- 45°
Worked answer
90°
If arcsin x = α then sinα = x, so cos(90° - α) = x too. The angle 90° - α lies between 0° and 180°, which is exactly the arccosine range, so arccos x = 90° - α. Adding gives α + (90° - α) = 90°, whatever x is.
Problem 6
Evaluate arctan(tan 135°).
- -45°
- 135°
- 45°
- -135°
- 225°
Worked answer
-45°
tan 135° = -1, since the reference angle is 45° and tangent is negative in quadrant II. Arctangent returns an angle strictly between -90° and 90°. The only one there with tangent -1 is -45°.
Problem 7
Evaluate cos(arctan 2).
- √55
- 2√55
- 12
- 2
- √5
Worked answer
√55
Let θ = arctan 2, so opposite = 2 and adjacent = 1. The hypotenuse is √12+22 = √5. So cosθ = 1√5 = √55, positive because arctangent returns angles where cosine is positive.
Problem 8
For what value of x is arcsin x = -30°?
- -12
- 12
- -√32
- -30
- √22
Worked answer
-12
The equation says the angle whose sine is x equals -30°. So x = sin(-30°). Sine is odd, so that is -sin 30° = -12. It lies inside [-1,1], as it must.