1. Given that sin x = 5/12 and 0< and = x < and = 90 degree
find the values of cos 2 x and cos 4 x .
2. Given that sin A and cos A = 3/4 ,find the values
a. sin 2 A
b. sin ^3 A + cos ^3 A
c. sin ^4 A + cos ^4 A
3. If 36 cos ^2 x - 3 sin x = 31 and 90 degree < x < 180 degree find the value of 2x .
Please come and help!!
2006-11-16 3:24 am
回答 (2)
2006-11-16 3:37 am
✔ 最佳答案
1.cos2x=1-2sinx^2 = 1 - 2 * (5/12)^2 = 1 - (50 / 144) = 47/72
cos4x = 2(cos2x)^2 - 1 = 2 * (47/72)^2 - 1 = -383/2592
2a
sin2A = 2sinAcosA=2*3/4*3/4=9/8
2b
(sinA)^3 + (cosA)^3 = (3/4)^3 * 2 = 27/32
2c
(sinA)^4 + (cosA)^4 = (3/4)^4 * 2 = 81 / 128
3
36 cos ^2 x - 3 sin x = 31
36(1-sin^2x) -3sinx=31
36 - 36sin^2x - 3sinx = 31
36sin^2x + 3sinx -5 = 0
(12sinx + 5)(3sinx -1) = 0
sinx = -5/12 or sinx = 1/3
Since 90 degree < x < 180 degree,
sinx = -5/12 rejected.
sinx = 1/3
x = 160.53degree
2x = 312.06degree
2006-11-25 5:59 am
1. 0< and = x < and = 90 degree implies cos x is +ve.
(cos x)^2 + (sin x)^2 = 1
(cos x)^2 = 1-(sin x)^2 = 1- 25/144 = 119 / 144
cos 2x = 2(cos x)^2 -1
= 2 (119 / 144) -1 = 47/72
cos 4x = 2(cos 2x)^2 -1
2(47/72)^2 -1 = -383/2592
(cos x)^2 + (sin x)^2 = 1
(cos x)^2 = 1-(sin x)^2 = 1- 25/144 = 119 / 144
cos 2x = 2(cos x)^2 -1
= 2 (119 / 144) -1 = 47/72
cos 4x = 2(cos 2x)^2 -1
2(47/72)^2 -1 = -383/2592
參考: me
收錄日期: 2021-04-12 22:57:03
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