標題:

Trigonometry2

發問:

Prove the following: 1. cos^2 42 + cos^2 51 + cos^2 69 + cos^2 78 = 1 + cos^2 72 + cos^2 81 2. sin^2 5 + sin^2 55 + sin^2 65 = 3/2

最佳解答:

1. LHS= cos^2 42 + cos^2 51 + cos^2 69 + cos^2 78= cos^2 (60-18) + cos^2 (60-9) + cos^2 (60+9) + cos^2 (60+18)= (cos 60 cos 18 + sin 60 sin 18)^2 + (cos 60 cos 9 + sin 60 sin 9)^2 + (cos 60 cos 9 - sin 60 sin 9)^2 + (cos 60 cos 18 - sin 60 sin 18)^2= 2 cos^2 60 cos^2 18 + 2 sin^2 60 sin^2 18 + 2 cos^2 60 cos^2 9 + 2 sin^2 60 sin^2 9= (1/2) cos^2 18 + (3/2) sin^2 18 + (1/2) cos^2 9 + (3/2) sin^2 9= (1/2) sin^2 72 + (3/2) cos^2 72 + (1/2) sin^2 81 + (3/2) cos^2 81= 1/2 - (1/2) cos^2 72 + (3/2) cos^2 72 + 1/2 - (1/2) cos^2 81 + (3/2) cos^2 81= 1 + cos^2 72 + cos^2 81= RHS 2. LHS= sin^2 5 + sin^2 55 + sin^2 65= sin^2 5 + sin^2 (60 - 5) + sin^2 (60 + 5)= sin^2 5 + (sin 60 cos 5 - cos 60 sin 5)^2 + (sin 60 cos 5 + cos 60 sin 5)^2= sin^2 5 + 2 sin^2 60 cos^2 5 + 2 cos^2 60 sin^2 5= sin^2 5 + (3/2) cos^2 5 + (1/2) sin^2 5= (3/2) sin^2 5 + (3/2) cos^2 5= 3/2= RHS

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