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26.6 Let
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ABC be any triangle. Using the notation in Fig. 26-10, prove sin B sin C sin A = = a b c law of sines
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(Here, sin A is the sine of CAB, and similarly for sin B and sin C.)
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Fig. 26-10
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Let D be the foot of the perpendicular from A to side BC. Let h = AD. Then, sin B = and so area of ABC = 1 1 1 (base height) = ah = ac sin B 2 2 2 h AD = c AB or h = c sin B
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SINE AND COSINE FUNCTIONS
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(Check that this is also correct when B is obtuse.) Similarly, area = Hence 1 1 1 ac sin B = bc sin A = ab sin C 2 2 2 and division by 1 abc gives the law of sines. 2 1 bc sin A 2 area = 1 ab sin C 2
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Supplementary Problems
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26.7 Evaluate: (a) sin 4 /3 ; (b) cos 11 /6 ; (c) sin 240 ; (d) cos 315 ; (e) sin 75 ; (f ) sin 15 ; (g) sin (11 /12); (h) cos 71 (i) sin 12 . 26.8 If is acute and cos = 1 , evaluate: 9 (a) sin ; (b) sin 2 ; (c) cos 2 ; (d) cos ( /2); (e) sin ( /2). (a) cos ; (b) sin 2 ; (c) cos 2 ; (d) cos ( /2); (e) sin( /2).
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26.9 If is in the third quadrant and sin = 2 , evaluate: 5 26.10 In ABC, AB = 5, AC = 7, and BC = 6. Find cos B.
26.11 In Fig. 26-11, D is a point on side BC of law of cosines twice.]
ABC such that AB = 2, AD = 5, BD = 4, and DC = 3. Find AC. [Hint: Use the
Fig. 26-11
26.12 Prove the following identities: (a) cos2 5 = (1 sin 5 )(1 + sin 5 ) (c) (sin + cos )2 = 1 + sin 2 (b)
Fig. 26-12
cos 1 + cos 2 = sin 2 sin (d) cos4 sin4 = cos2 sin2
SINE AND COSINE FUNCTIONS
[CHAP. 26
26.13 For each of the following, either prove that it is an identity or give an example in which it is false: cos 1 sin + = (a) sin 4 = 4 sin cos (b) 1 + cos sin sin (c) sin4 + cos4 = 1 (e) cos sin 2 + = sin cos sin 2 cos sin2 = cos cos2 1 + cos 3 3 (f ) 2 sin cos = sin 3 2 2 (d)
26.14 Fill in all details of the following proof of the identity cos (u v) = cos u cos v + sin u sin v Consider the case where 0 v < u < v + (see Fig. 26-12). By the law of cosines, BC = 12 + 12 2(1)(1) cos  BOC (cos u cos v)2 + (sin u sin v)2 = 2 2 cos (u v) cos2 u 2 cos u cos v + cos2 v + sin2 u 2 sin u sin v + sin2 v = 2 2 cos (u v) (cos2 u + sin2 u) + (cos2 v + sin2 v) 2(cos u cos v + sin u sin v) = 2 2 cos (u v) 1 + 1 2(cos u cos v + sin u sin v) = 2 2 cos(u v) cos u cos v + sin u sin v = cos (u v) Verify that all other cases can be derived from the case just considered. 26.15 Prove the following identities: (a) cos + = sin 2 (c) cos ( + ) = cos = cos 2 (d) sin ( + ) = sin (b) sin +
1 2 3 26.16 Show: (a) sin = cos = (b) sin = cos = (c) sin = cos = 4 4 2 6 3 2 3 6 2 [Hint: (a) Take an isosceles right triangle OGB with right angle at G and let b = OG BG and c = OB. = 2 b 2 1 b2 1 . (b) Take an equilateral triangle = 2 = . Hence, sin = = Then, c2 = b2 + b2 = 2b2 . So, sin2 = 4 c 2 4 2 c 2 1. ABC of side 1 and let AD be the altitude from A to the midpoint D of BC. Then BD = 2 1 BD = 2 = . By Theorem 26.7, sin = cos radians, cos = = cos . 3 3 1 2 6 2 6 3 AB 2 = 1 cos2 = 1 1 = 3 . Hence, sin = 3 and cos = sin by Theorem 26.7.] (c) sin 3 3 4 4 3 2 6 3 Since  ABD contains 26.17 Derive the other parts of Theorem 26.6 from part (b), which was proved in Problem 26.14. [Hint: First note that cos = sin follows from part (b). The formula for cos (u + v) follows by applying part (b) to cos (u ( v)) and 2 using the identities cos ( v) = cos v and sin ( v) = sin v. The formula for sin (u + v) follows by applying part (b) to u v , and then the formula for sin (u v) follows from the formula for sin (u + v) by replacing v by v.] cos 2
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