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分类: 作业答案 • 2022-05-07 12:34:41

问题描述：

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数学人气：258 ℃时间：2020-09-03 11:43:41

优质解答

cos(a+π/4)=cos(a)*cos(π/4) - sin(a)*sin(π/4) =(1/2^0.5)*(cos( a) - sin(a)) = 4/5
cos( a) - sin(a) = (4/5)*2^0.5.(1)
(cos(a+π/4))^2 +(sin(a + π/4))^2 = 1 -> (sin(a+π/4))^2 = 1 - 16/25 = 9/25
0 sin(a+π/4) = 3/5
sin(a+π/4) = sin(a )*cos(π/4) + cos(a)*sin(π/4) =(1/2^0.5)(cos(a) + sin(a))= 3/5
cos(a) + sin(a) = (3/5)*2^0.5.(2)
(1) + (2) -> 2cos(a) = (7/5)*2^0.5 -> cos(a) = (7/10)*2^0.5
(2) - (1) -> 2sin(a) = (-1/5)*2^0.5 -> sin(a) = (-1/10)*2^0.5
tan(a)= sin(a)/cos(a) = -1/7
补充:题目应该是cos(a+π/4) = -4/5,则解为-7

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答

cos(a+π/4)=cos(a)*cos(π/4) - sin(a)*sin(π/4) =(1/2^0.5)*(cos( a) - sin(a)) = 4/5
cos( a) - sin(a) = (4/5)*2^0.5.(1)
(cos(a+π/4))^2 +(sin(a + π/4))^2 = 1 -> (sin(a+π/4))^2 = 1 - 16/25 = 9/25
0 sin(a+π/4) = 3/5
sin(a+π/4) = sin(a )*cos(π/4) + cos(a)*sin(π/4) =(1/2^0.5)(cos(a) + sin(a))= 3/5
cos(a) + sin(a) = (3/5)*2^0.5.(2)
(1) + (2) -> 2cos(a) = (7/5)*2^0.5 -> cos(a) = (7/10)*2^0.5
(2) - (1) -> 2sin(a) = (-1/5)*2^0.5 -> sin(a) = (-1/10)*2^0.5
tan(a)= sin(a)/cos(a) = -1/7
补充:题目应该是cos(a+π/4) = -4/5,则解为-7

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