Inverse Trig Functions
A free Trigonometry lesson from the “Inverse Trig and Equations” unit, with a worked example and practice problems including step-by-step solutions.
Because sine, cosine, and tangent repeat their values, each function is restricted to a single interval before it can be inverted, so its inverse returns exactly one angle. The inverse functions return that one principal-value angle: arcsin in [-90, 90] deg, arccos in [0, 180] deg, and arctan in (-90, 90) deg. Evaluating an inverse trig function means asking "which angle in the principal range has this sine, cosine, or tangent?" rather than solving an equation with infinitely many solutions.
What you'll learn
- Explain why each inverse trig function uses a restricted principal-value range: arcsin in [-90, 90] deg, arccos in [0, 180] deg, and arctan in (-90, 90) deg.
- Evaluate arcsin, arccos, and arctan at standard inputs and report the single principal-value angle in degrees or radians.
- Identify the domain and range of each inverse trig function and explain why an output like arccos(-1/2) is positive while arcsin(-sqrt(2)/2) is negative.
Worked example
Problem. Evaluate arccos(-1/2), giving the answer in degrees.
- arccos asks: which angle in the principal range [0, 180] deg has cosine equal to -1/2?
- Cosine is -1/2 at a reference angle of 60 deg; since the value is negative, the angle is in the second quadrant: 180 - 60 = 120 deg.
- 120 deg is inside [0, 180], so arccos(-1/2) = 120 deg = 2pi/3.
Answer: 120 degrees (2pi/3)
Practice problems
1. Evaluate arcsin(1/2). Give the answer in degrees.
Show solution
- arcsin asks which angle in [-90, 90] deg has sine 1/2.
- sin(30 deg) = 1/2 and 30 deg is in the range.
- So arcsin(1/2) = 30 deg = pi/6.
Answer: 30 degrees (pi/6)
2. What is the principal-value range of arccos?
Choices: [-90 deg, 90 deg] · [0 deg, 180 deg] · (-90 deg, 90 deg) · [0 deg, 360 deg]
Show solution
- Cosine is restricted to [0, 180] deg so that it is one-to-one before inverting.
- Every arccos output therefore lies in [0, 180] deg.
- The correct range is [0 deg, 180 deg].
Answer: [0 deg, 180 deg]
3. Evaluate arctan(1). Give the answer in degrees.
Show solution
- arctan asks which angle in (-90, 90) deg has tangent 1.
- tan(45 deg) = 1 and 45 deg is in the range.
- So arctan(1) = 45 deg = pi/4.
Answer: 45 degrees (pi/4)
4. Evaluate arccos(-1/2). Give the answer in degrees.
Show solution
- Find the angle in [0, 180] deg with cosine -1/2.
- Reference angle 60 deg, negative cosine means second quadrant: 180 - 60 = 120 deg.
- So arccos(-1/2) = 120 deg = 2pi/3.
Answer: 120 degrees (2pi/3)
5. Evaluate arcsin(-sqrt(2)/2). Give the answer in degrees.
Show solution
- Find the angle in [-90, 90] deg with sine -sqrt(2)/2.
- Reference angle 45 deg, negative sine means a negative angle: -45 deg.
- So arcsin(-sqrt(2)/2) = -45 deg = -pi/4.
Answer: -45 degrees (-pi/4)
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