１ｄ三角比既問題

sin 30 = 0.5; sin 45 = 1/[sqrt 2] ; sin 60 = [sqrt 3]/2
cos 30 = [sqrt 3]/2 ; cos 45 = 1/[sqrt 2] ; cos 60 = 0.5
tan 30 = 1/[sqrt 3] ; tan 45 = 1; tan 60 = [sqrt 3]

4.2 sin^2 30度＋cos^2 45度* tan^2 60度
=2(0.5)^2 + (1/[sqrt 2])^2 * ([sqrt 3])^2
=2

6.1/cos45度+1/sin45度
=1/(1/[sqrt 2]) + 1/(1/[sqrt 2])
=2[sqrt 2]

10.4sin 60度*cos30度=3tan (feator)
==> 3tan(feator) = 4*[sqrt3]/2*[sqrt 3]/2 = 1
==> tan(feator) = 1
==> (feator) = 180n + 45 where n = 0,1,2,3,...
==> (feator) = 45, 225, 405, 585,...

18.tan(feator)/2*tan30度*cos30度=cos60度
==> tan(feator) = 0.5*2/(1/[sqrt 3])/([sqrt3]/2)
==> tan(feator) = 2
==> feator = 180n + tan^-1(2) = 63.4, 243.4, 423.4.

3.已知cos(feator)=9/41,求sin(feator)及tan(feator)既值。
sin(feator) = sqrt(1-cos(feator)^2) = 40/41
tan(feator) = sin(feator)/cos(feator) = 40/9

6.已知tan(feator)=5,求sin(feator)及cos(feator)的值。
tan(feator) = 5/1
sin(feator) = 5/[sqrt 5^2+1^2] = 5/[sqrt 26]
cos(feator) = sin(feator)/tan(feator) = 1/[sqrt 26]

7.已知sin(feator)=9/41,求tan(feator)/(1+tan(feator))的值。
tan(feator) = 9/[sqrt 41^2-9^2] = 9/40
tan(feator)/(1+tan(feator)) = (9/40)(1+9/40) = 9/49

8.已知tan(o)3.求sin(o)+2cos(o)/2sin(o)+cos(o)的值。
sin(o) = 3/[sqrt 3^2 + 1^2] = 3/[sqrt 10]
cos(o) = 1/[sqrt 3^2 + 1^2] = 1/[sqrt 10]
sin(o)+2cos(o)/2sin(o)+cos(o)
= 2/[sqrt 10] + 2*1/[sqrt 10]*3/[sqrt 10] + 1/[sqrt 10]
= 3/[sqrt 10] + 3/5

10.已知sin(feator)=8/17,求3tan(feator)-2/2cos(feator)-sin(feator)的值。
tan(feator) = 8/[sqrt 17^2 - 8^2] = 8/15
cos(feator) = sin(feator)/tan(feator) = 15/17
3tan(feator)-2/2cos(feator)-sin(feator)
= 3*15/17 -2/(2*15/17) - 8/17
= 266/255