Interpret principal inverse trigonometric values
Read inverse notation as a function value, distinguish it from a reciprocal and explain why it gives only one selected angle.
Section 1.5 · Trigonometry
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. State angle units and find every solution in the supplied interval; general forms of solution are not required in section 1.5. Written practice is self-checked.
Before you begin
- Understand a one-to-one function and an inverse.
- Know the signs and exact values of basic trigonometric ratios.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Can a real sin⁻¹(1.2) be evaluated?
What you will learn
- Use sin⁻¹, cos⁻¹ and tan⁻¹ to denote principal inverse values.
- State their principal ranges and distinguish one inverse value from all solutions of an equation.
Restrict the original function to make an inverse
Sine, cosine and tangent repeat, so an unrestricted ratio cannot identify a unique input angle. Principal inverse values choose branches where the original function is one-to-one. In radians, sin⁻¹ has output range [−π/2, π/2], cos⁻¹ has [0, π], and tan⁻¹ has (−π/2, π/2). The degree equivalents are [−90°, 90°], [0°, 180°] and (−90°, 90°).
Real sin⁻¹ and cos⁻¹ accept inputs only in [−1, 1]; tan⁻¹ accepts every real input. For example sin⁻¹(−√3/2) = −π/3, cos⁻¹(−1/2) = 2π/3 and tan⁻¹(−1) = −π/4. Select calculator mode for the intended angle unit, and do not treat an out-of-range sine input as a calculator fault.
The superscript −1 denotes an inverse, not a reciprocal
sin⁻¹(1/2) is the principal angle π/6; 1/sin(π/6) is the reciprocal number 2. They are different operations with different inputs and outputs. Likewise, sin⁻¹(sin 150°) = 30°, because 150° is outside sine's principal inverse range. Inverting an output recovers the original input only when that input belongs to the selected branch.
If sin θ = 1/2 for 0° ≤ θ ≤ 180°, the inverse key gives 30° but the graph also gives 150°. The inverse function returns one principal value; solving the equation requires every value in the supplied interval. Here the principal branches are the useful ideas; specialised inverse-function identities are outside this introductory section.
Pause and explain
What does sin⁻¹(1/2) return in degrees?
Worked example
Find sin⁻¹(−√3/2), cos⁻¹(−1/2) and tan⁻¹(−1) in radians.
Show the worked solution
- The sine inverse range is [−π/2, π/2], so choose −π/3.
- The cosine inverse range is [0, π], so choose 2π/3.
- The tangent inverse range is (−π/2, π/2), so choose −π/4; each selected angle produces the supplied ratio.
Answer −π/3, 2π/3 and −π/4 respectively.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Evaluate sin⁻¹(1/2) and cos⁻¹(1/2) exactly in radians.
Give me a hint
Both use the ratio 1/2, but their functions and principal branches differ.
Compare my reasoning
- Within sine's principal interval, sin(π/6) = 1/2.
- Within cosine's principal interval, cos(π/3) = 1/2.
- Both values belong to their appropriate output ranges and can be checked in the original functions.
sin⁻¹(1/2) = π/6; cos⁻¹(1/2) = π/3.
Look for these in your work
- Choose the original function's special angle.
- Check the principal interval and retain exact radian form.
Evaluate cos⁻¹(−√2/2) and sin⁻¹(−√2/2) in degrees.
Give me a hint
Use the 45° reference angle but choose a value in each inverse's own output range.
Compare my reasoning
- For cosine, choose the quadrant-II angle 135° within [0°, 180°].
- For sine, choose −45° within [−90°, 90°].
- The same ratio does not require the same principal angle; substitution verifies both.
cos⁻¹(−√2/2) = 135°; sin⁻¹(−√2/2) = −45°.
Look for these in your work
- Use different principal ranges for sine and cosine.
- Verify the sign of the ratio at each selected angle.
A sensor gives sin θ = 1/2 with 0° ≤ θ ≤ 180°. A learner records only θ = sin⁻¹(1/2) = 30°. Explain and complete the answer.
Give me a hint
The supplied angle interval is wider than the branch chosen by the inverse key.
Compare my reasoning
- The inverse key correctly returns the principal value 30°.
- Sine is also positive with the same magnitude at 180° − 30° = 150°.
- Both angles lie in the sensor's interval and give 1/2; no other angle there does.
θ = 30° or 150°; the principal inverse alone misses the second permitted angle.
Look for these in your work
- Distinguish a correct principal value from a complete solution set.
- Use the given physical interval to retain both possible angles.
Common mistakes
- Inverse notation does not mean the reciprocal of a trigonometric value.
- The inverse key returns one principal angle, rather than every equation solution.
- Sine and cosine inverse inputs must be between −1 and 1.
What can you explain now?
Evaluate sin⁻¹(sin 150°) in degrees and explain the result.
Compare with the explanation
30°, because sine's principal inverse range is [−90°, 90°].
sin 150° = 1/2, whose principal sine inverse is 30°, not the original angle outside that branch.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
State each inverse's input and principal output ranges. Give an example where applying sin⁻¹ to a sine value does not recover the original angle.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.