1.Solve the following equations, where 0°≦x＜360°
a) cos x =sin100°
b) sin(x + 100°)=cos120°
2.a) If 23516=360n+x, where n is a positive integer and 0<x<360, find n and x.
b) Hence express tan23516 in trigonometric ratio of an acute angle.
3.a)Prove that (2-x)/(3+x) = 5/(3+x)-1
b)Hence find the maximum and minimum values of (2-siny)/(3+siny)
4)Given that sinx-cosx=m, express each of the following in terms of m.
a)sinxcosx
b)sin^(2)x-sin^(4)x
trigonometry
2012-08-27 8:14 am
回答 (2)
2012-08-27 8:52 am
✔ 最佳答案
1a)cosx=sin100°=cos10°
x=10° or 350°
1b)
sin(x+100°)=cos120°=sin(330°)
x+100°=210° or 330°
x = 110° or 230°
2a)
23516=360n+x
sin(23516)=sin(360n+x)
sin(360*65+116)=sin(360n+x)
n=65 and x=116
2b)
tan23516° = tan65°
3a)
L.H.S.: (2-x)/(3+x)
= (-3-x +5)/(3+x)
= 5/(3+x) - 1 =R.H.S.
3b)
(2-siny)/(3+siny) = 5/(3+siny) -1
then , -1 =< siny =< 1
3/2 >= 5/(3+siny) -1 >=1/4
Hence , the max. is 3/2 , min. is 1/4
4a)
m^2 = sin^2(x)+cos^2(x)-2sinxcosx
= 1 -2sinxcosx
sinxcosx = (1+m)(1-m)/2
4b)
sin^2(x) - sin^4(x) = sin^2(x)*cos^2(x)
= ((1+m)(1-m)/2)^2
= ((1-m^2)^2)/4
參考: 神風亨
2012-08-27 9:27 am
2.
a)
23516 ÷ 360 = 65 ...... remainder= 116
Hence, 23516 = 360*65 + 116
n = 65 and x = 116
b)
tan23516°
= tan(360°*65 + 116°)
= tan116°
= tan(180° - 64°)
= -tan64°
a)
23516 ÷ 360 = 65 ...... remainder= 116
Hence, 23516 = 360*65 + 116
n = 65 and x = 116
b)
tan23516°
= tan(360°*65 + 116°)
= tan116°
= tan(180° - 64°)
= -tan64°
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