Examkin

Soru

Zorluk: OrtaInverse 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)?

Cevap: 180 degrees

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Evaluating inverse trigonometric functions within their standard principal value ranges.
Bu soruyu puanla

Konular

Inverse Trigonometric Functions

Bu konudan daha fazla soru