Let’s solve each inverse trig function one by one. Remember:
- sin⁻¹(x) gives you the angle whose sine is x, and the answer must be between -π/2 and π/2 (or -90° to 90°).
- cos⁻¹(x) gives you the angle whose cosine is x, and the answer must be between 0 and π (or 0° to 180°).
- tan⁻¹(x) gives you the angle whose tangent is x, and the answer must be between -π/2 and π/2 (or -90° to 90°).
We’ll use degrees for simplicity unless told otherwise — but note that in higher math, radians are standard. Since this looks like a basic homework problem, we’ll give answers in both degrees and radians so you’re covered.
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1. sin⁻¹(-½)
What angle has a sine of -½?
We know sin(30°) = ½ → so sin(-30°) = -½.
And -30° is within the allowed range for sin⁻¹ (-90° to 90°).
In radians: -30° = -π/6.
✔ Answer:
-30° or -π/6
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2. sin⁻¹(-1)
What angle has a sine of -1?
sin(-90°) = -1.
-90° is within the allowed range.
In radians: -90° = -π/2.
✔ Answer:
-90° or -π/2
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3. cos⁻¹(0)
What angle has a cosine of 0?
cos(90°) = 0.
90° is within the allowed range for cos⁻¹ (0° to 180°).
In radians: 90° = π/2.
✔ Answer:
90° or π/2
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4. cos⁻¹(-√3/2)
What angle has a cosine of -√3/2?
We know cos(30°) = √3/2 → so cosine is negative in quadrant II.
So the angle is 180° - 30° = 150°.
Check: cos(150°) = -√3/2 ✔️
150° is within 0° to 180° → valid.
In radians: 150° = 5π/6.
✔ Answer:
150° or 5π/6
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5. tan⁻¹(√3)
What angle has a tangent of √3?
tan(60°) = √3.
60° is within the allowed range for tan⁻¹ (-90° to 90°).
In radians: 60° = π/3.
✔ Answer:
60° or π/3
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6. tan⁻¹(-√3/3)
First, simplify: √3/3 = 1/√3 → so we want tan⁻¹(-1/√3)
We know tan(30°) = 1/√3 → so tan(-30°) = -1/√3
-30° is within the allowed range.
In radians: -30° = -π/6.
✔ Answer:
-30° or -π/6
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Final Answer:
sin⁻¹(-½) = -30° or -π/6
sin⁻¹(-1) = -90° or -π/2
cos⁻¹(0) = 90° or π/2
cos⁻¹(-√3/2) = 150° or 5π/6
tan⁻¹(√3) = 60° or π/3
tan⁻¹(-√3/3) = -30° or -π/6
Parent Tip: Review the logic above to help your child master the concept of inverse trigonometric.
