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Math  /  Trigonometry

QuestionSections 4.7+4.84.7+4.8
Show all work! (1) Find the exact value of each expression. state if undefined. a) arccos⁡(12)\arccos \left(\frac{1}{2}\right) b) arcsin⁡(4)\arcsin (4) c) sin⁡(arcsin⁡(−12))\sin \left(\arcsin \left(-\frac{1}{2}\right)\right) d) tan⁡(arccos⁡(37)) sketcl this on the  coordinate plane. \tan \left(\arccos \left(\frac{3}{7}\right)\right) \quad \begin{array}{l}\text { sketcl this on the } \\ \text { coordinate plane. }\end{array} (2) Solve the problem. Use exact values (leave in terms of a trig function. Aski slope is 52 ft long and the angle A ski slope is from the ground to the summit is 42∘42^{\circ}. How high is the summit? (Draw your best ski slope and mountain) "̈

Studdy Solution

STEP 1

1. The range of the inverse trigonometric functions is assumed to be the principal value range.
2. For arccos⁡\arccos, the range is [0,π][0, \pi].
3. For arcsin⁡\arcsin, the range is [−π2,π2][- \frac{\pi}{2}, \frac{\pi}{2}].
4. The trigonometric functions are defined for real numbers unless otherwise specified.
5. The ski slope problem involves a right triangle with the slope as the hypotenuse.

STEP 2

1. Evaluate each inverse trigonometric expression.
2. Solve the ski slope problem using trigonometry.

STEP 3

Evaluate arccos⁡(12)\arccos\left(\frac{1}{2}\right).
arccos⁡(12) \arccos\left(\frac{1}{2}\right) corresponds to the angle whose cosine is 12\frac{1}{2}. In the principal range [0,π][0, \pi], this angle is:
π3 \frac{\pi}{3}

STEP 4

Evaluate arcsin⁡(4)\arcsin(4).
The sine function has a range of [−1,1][-1, 1]. Since 44 is outside this range, arcsin⁡(4)\arcsin(4) is:
Undefined

STEP 5

Evaluate sin⁡(arcsin⁡(−12))\sin\left(\arcsin\left(-\frac{1}{2}\right)\right).
The expression arcsin⁡(−12)\arcsin\left(-\frac{1}{2}\right) gives the angle whose sine is −12-\frac{1}{2}. In the principal range [−π2,π2][- \frac{\pi}{2}, \frac{\pi}{2}], this angle is:
−π6 -\frac{\pi}{6}
Thus, sin⁡(arcsin⁡(−12))\sin\left(\arcsin\left(-\frac{1}{2}\right)\right) is:
−12 -\frac{1}{2}

STEP 6

Evaluate tan⁡(arccos⁡(37))\tan\left(\arccos\left(\frac{3}{7}\right)\right).
Sketch a right triangle where arccos⁡(37)\arccos\left(\frac{3}{7}\right) is the angle. The adjacent side is 33, and the hypotenuse is 77. Use the Pythagorean theorem to find the opposite side:
Opposite side=72−32=49−9=40=210 \text{Opposite side} = \sqrt{7^2 - 3^2} = \sqrt{49 - 9} = \sqrt{40} = 2\sqrt{10}
Thus, tan⁡(arccos⁡(37))\tan\left(\arccos\left(\frac{3}{7}\right)\right) is:
OppositeAdjacent=2103 \frac{\text{Opposite}}{\text{Adjacent}} = \frac{2\sqrt{10}}{3}

STEP 7

Solve the ski slope problem.
Given: - Hypotenuse (slope) = 5252 ft - Angle = 42∘42^\circ
We need to find the height of the summit, which is the opposite side of the angle in a right triangle. Use the sine function:
sin⁡(42∘)=Height52 \sin(42^\circ) = \frac{\text{Height}}{52}
Solve for Height:
Height=52⋅sin⁡(42∘) \text{Height} = 52 \cdot \sin(42^\circ)
Since we need the exact value in terms of a trig function:
Height=52⋅sin⁡(42∘) \text{Height} = 52 \cdot \sin(42^\circ)
The solutions are: a) π3\frac{\pi}{3} b) Undefined c) −12-\frac{1}{2} d) 2103\frac{2\sqrt{10}}{3} 2) Height = 52⋅sin⁡(42∘)52 \cdot \sin(42^\circ)

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