sin, cos , tan

2007-03-12 1:41 am
化簡
1) 1/sin@ - sin@ / tan二次@
2) 若cos@=7/25, 求 (sin@ - 2cos@) / sin@ + 3cos@
證明是否恆等式
3) cos@/tan@ + sin@ = 1/sin@

回答 (1)

2007-03-12 2:04 am
✔ 最佳答案
1:
1/sin@-sin@/tan^2@=sin@/sin^2@-sin@/(sin^2@/cos^2@)
=sin@/sin^2@-sin@cos^2@/sin^2@
=(sin@-sin@cos^2@)/sin^2@
=sin@(1-cos^2@)/sin^2@
=sin@(sin^2@)/sin^2@
=sin@

2

cos@=7/25
斜邊=25 底邊=7 對邊=開方(25^2-7^2)=24

因cos@=7/25 (>0)
@ lies in quadrant I or IV.

When @ lies in quadrant I, tan@=24/7
when @ lies in quadrant IV, tan@=-24/7 (如果你是中三或以下可以不理會這裏)

(sin@-2cos@)/(sin@+3cos@)=[(sin@-2cos@)/cos@]/[(sin@+3cos@)/cos@]
=(tan@-2)/(tan@-3)
=(24/7-2)/(24/7-3) or = (-24/7-2)/(-24/7-3)
=10/3 or =38/45(如果你是中三或以下可以不理會這裏)

3:
L.H.S.=cos@/tan@+sin@
= cos@/(sin@/cos@)+sin@
=cos@(cos@)/sin@+sin^2@/sin@
=(cos^2@+sin^2@)/sin@
=1/sin@
=R.H.S
參考: me^^


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