sin , cos equation........

Ggg

2011-10-31 5:31 pm

1. Solve sin(4x)=cos(2x) for 0 degree to 180 degree. (Tips. there are 4 answers)

2. It is given that A,B.C are the interior angles of triangle ABC respectively, Also sin(A)^2 + sin(B)^2=1-cos(A+B)cos(A-B). Prove that sin(A)^2+sin(B)^2+sin(C)^2=2+2cos(A)cos(B)cos(C).

回答 (1)

用戶25975

2011-10-31 6:16 pm

✔ 最佳答案

1) sin 4x = cos 2x

2 sin 2x cos 2x = cos 2x

cos 2x (2 sin 2x - 1) = 0

cos 2x = 0 or sin 2x = 1/2

2x = 90, 270, 30 or 150

x = 45, 135, 15 or 75

2) sin2 A + sin2 B + sin2 C

= 1 - cos (A + B) cos (A - B) + 1 - cos2 C

= 2 - cos (180 - C) cos (A - B) - cos2 C

= 2 + cos (A - B) cos C - cos2 C

= 2 + cos C [cos (A - B) - cos C]

= 2 + cos C x 2 sin [(A - B + C)/2] sin [(C - A + B)/2]

= 2 + 2 cos C sin [(180 - 2B)/2] sin [(180 - 2A)/2]

= 2 + 2 cos C sin (90 - B) sin (90 - A)

= 2 + 2 cos A cos B cos C

參考: 原創答案

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