le b. 51. Suppose that tan 0 = p/q, where p and q are positive ani 0° < 0 < 90°. Show that p sin0 q cos O p² – q° amtutmat p sin 0 + q cos 0 p² + q² +g² 52. This exercise shows how to obtain radical expressions for sin 15° and cos 15°. In the figure, assume that AB = BD = 2. %3D niwe 30° (a) In the right triangle BCD, note that DC = 1 because DC is opposite the 30° angle and BD = 2. Use the Pythagorean theorem to show that BC = V3. (b) Use the Pythagorean theorem to show that AD = 2\2 + V3. (c) Show that the expression for AD in part (b) is equal to V6 + V2. Hint: Two nonnegative quantities are 54. Thi %3D sin equal if and only if their squares are equal. (d) Explain why ZBAD = LBDA. (e) According to a theorem from geometry, an exterior angle of a triangle is equal to the sum of the two nonadjacent interior angles. Apply this to AABD with exterior angle DBC = 30°, and show that ZBAD = 15°. (f) Using the figure and the values that you have obtained for the lengths, conclude that (a (b 1 2 + V3 sin 15° cos 15° = Võ + V2 (c (g) Rationalize the denominators in part (f) to obtain Võ + V2 sin 15° cos 15° 4 4. (h) Use your calculator to check the results in part (g). 3. The following foure chows a regular eleven-sided polygon

Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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Number 52

le
b.
51. Suppose that tan 0 = p/q, where p and q are positive ani
0° < 0 < 90°. Show that
p sin0
q cos O
p² – q°
amtutmat
p sin 0 + q cos 0
p² + q²
+g²
52. This exercise shows how to obtain radical expressions
for sin 15° and cos 15°. In the figure, assume that
AB = BD = 2.
%3D
niwe
30°
(a) In the right triangle BCD, note that DC = 1
because DC is opposite the 30° angle and BD = 2.
Use the Pythagorean theorem to show that
BC = V3.
Transcribed Image Text:le b. 51. Suppose that tan 0 = p/q, where p and q are positive ani 0° < 0 < 90°. Show that p sin0 q cos O p² – q° amtutmat p sin 0 + q cos 0 p² + q² +g² 52. This exercise shows how to obtain radical expressions for sin 15° and cos 15°. In the figure, assume that AB = BD = 2. %3D niwe 30° (a) In the right triangle BCD, note that DC = 1 because DC is opposite the 30° angle and BD = 2. Use the Pythagorean theorem to show that BC = V3.
(b) Use the Pythagorean theorem to show that
AD = 2\2 + V3.
(c) Show that the expression for AD in part (b) is equal
to V6 + V2. Hint: Two nonnegative quantities are
54. Thi
%3D
sin
equal if and only if their squares are equal.
(d) Explain why ZBAD = LBDA.
(e) According to a theorem from geometry, an exterior
angle of a triangle is equal to the sum of the two
nonadjacent interior angles. Apply this to AABD
with exterior angle DBC = 30°, and show that
ZBAD = 15°.
(f) Using the figure and the values that you have obtained
for the lengths, conclude that
(a
(b
1
2 + V3
sin 15°
cos 15° =
Võ + V2
(c
(g) Rationalize the denominators in part (f) to obtain
Võ + V2
sin 15°
cos 15°
4
4.
(h) Use your calculator to check the results in part (g).
3. The following foure chows a regular eleven-sided polygon
Transcribed Image Text:(b) Use the Pythagorean theorem to show that AD = 2\2 + V3. (c) Show that the expression for AD in part (b) is equal to V6 + V2. Hint: Two nonnegative quantities are 54. Thi %3D sin equal if and only if their squares are equal. (d) Explain why ZBAD = LBDA. (e) According to a theorem from geometry, an exterior angle of a triangle is equal to the sum of the two nonadjacent interior angles. Apply this to AABD with exterior angle DBC = 30°, and show that ZBAD = 15°. (f) Using the figure and the values that you have obtained for the lengths, conclude that (a (b 1 2 + V3 sin 15° cos 15° = Võ + V2 (c (g) Rationalize the denominators in part (f) to obtain Võ + V2 sin 15° cos 15° 4 4. (h) Use your calculator to check the results in part (g). 3. The following foure chows a regular eleven-sided polygon
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