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Difficulty: MediumInverse Trigonometric Functions

What is the value, in degrees, of the expression arcsin⁡(32)+arccos⁡(−12)\arcsin\left(\frac{\sqrt{3}}{2}\right) + \arccos\left(-\frac{1}{2}\right)?

Answer: 180 degrees

Answer

The value of the expression is 180 degrees.
To evaluate arcsin⁡(32)+arccos⁡(−12)\arcsin\left(\frac{\sqrt{3}}{2}\right) + \arccos\left(-\frac{1}{2}\right), find the principal value for each term. The principal range for arcsin⁡\arcsin is [−90∘,90∘][-90^\circ, 90^\circ], so arcsin⁡(32)=60∘\arcsin\left(\frac{\sqrt{3}}{2}\right) = 60^\circ. The principal range for arccos⁡\arccos is [0∘,180∘][0^\circ, 180^\circ], so for a negative argument, the result lies in Quadrant II, giving arccos⁡(−12)=120∘\arccos\left(-\frac{1}{2}\right) = 120^\circ. Adding these values together gives 60∘+120∘=180∘60^\circ + 120^\circ = 180^\circ.

Step-by-Step Solution

1
Find the principal angle for arcsin⁡(32)\arcsin\left(\frac{\sqrt{3}}{2}\right) in degrees.
arcsin⁡(32)=60∘\arcsin\left(\frac{\sqrt{3}}{2}\right) = 60^\circ
The inverse sine function yields outputs restricted to the range [−90∘,90∘][-90^\circ, 90^\circ]. The angle in Quadrant I whose sine is 32\frac{\sqrt{3}}{2} is 60∘60^\circ.
2
Find the principal angle for arccos⁡(−12)\arccos\left(-\frac{1}{2}\right) in degrees.
arccos⁡(−12)=120∘\arccos\left(-\frac{1}{2}\right) = 120^\circ
The inverse cosine function yields outputs restricted to the range [0∘,180∘][0^\circ, 180^\circ]. For a negative input, the output must be in Quadrant II. The angle whose cosine is −12-\frac{1}{2} is 120∘120^\circ.
3
Sum the two evaluated angle measures.
60∘+120∘=180∘60^\circ + 120^\circ = 180^\circ
Perform standard addition on the two principal values.

Key Concept

Evaluating inverse trigonometric functions within their standard principal value ranges.
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Topics

Inverse Trigonometric Functions

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