標題:
數學 AP GP問題~~~pls help!!!
發問:
(sin 1)^2 + (sin 3)^2 + (sin 5)^2 +........+ (sin 87)^2 + (sin89)^2 = ? ^2 means 二次方 更新: the answer is 22.5 Why?
最佳解答:
From trigonometry knowledge, sin 1 = cos (90 - 1) i.e. = cos 89 sin 3 = cos (90 - 3) = cos 87 ... sin 89 = cos (90 -1) =cos 1 etc Therefore, you can "group" the 1st term with the last term, 2nd term with the 2nd last term etc etc etc [(sin 1)^2 + (sin89)^2] + [(sin 3)^2 + (sin 87)^2] + ..... (sin45)^2 <--- ( (sin45)^2 will be the odd one out without a "pair") use equations in AP for the "numbers" in the sin function, (actually not necessary to use AP, it can be easily figure out) a = 1, d = 2 I = a + (n-1)d (I = nth term) 89 = 1 + (n-1)2 n = 45 Therefore, there are 45 terms. (22 "pairs" and 1 odd one) using sin^2 X + cos^2 X =1 [(sin 1)^2 + (sin89)^2] + [(sin 3)^2 + (sin 87)^2] + ..... (sin45)^2 will become 1 x 22 + (1/√2)^2 (because [(sin 1)^2 + (sin89)^2] = [(sin 1)^2 + (cos1)^2] = 1) = 22 + 1/2 = 22.5
[ (sin 1)^2 + (sin89)^2 ]x20=20
數學 AP GP問題~~~pls help!!!
發問:
(sin 1)^2 + (sin 3)^2 + (sin 5)^2 +........+ (sin 87)^2 + (sin89)^2 = ? ^2 means 二次方 更新: the answer is 22.5 Why?
最佳解答:
From trigonometry knowledge, sin 1 = cos (90 - 1) i.e. = cos 89 sin 3 = cos (90 - 3) = cos 87 ... sin 89 = cos (90 -1) =cos 1 etc Therefore, you can "group" the 1st term with the last term, 2nd term with the 2nd last term etc etc etc [(sin 1)^2 + (sin89)^2] + [(sin 3)^2 + (sin 87)^2] + ..... (sin45)^2 <--- ( (sin45)^2 will be the odd one out without a "pair") use equations in AP for the "numbers" in the sin function, (actually not necessary to use AP, it can be easily figure out) a = 1, d = 2 I = a + (n-1)d (I = nth term) 89 = 1 + (n-1)2 n = 45 Therefore, there are 45 terms. (22 "pairs" and 1 odd one) using sin^2 X + cos^2 X =1 [(sin 1)^2 + (sin89)^2] + [(sin 3)^2 + (sin 87)^2] + ..... (sin45)^2 will become 1 x 22 + (1/√2)^2 (because [(sin 1)^2 + (sin89)^2] = [(sin 1)^2 + (cos1)^2] = 1) = 22 + 1/2 = 22.5
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其他解答:[ (sin 1)^2 + (sin89)^2 ]x20=20
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