1. Given that cos(42-Y)tan(42-Y) = cos(2Y+18), where Y is an acute angle.
a) Find x by using the trigonometric identities.
b) By using the result obtained in (a), find the value of 3cosYsin(15+Y)tan 2Y
2. a) Prove that sin^2 Y - cos^2 Y = 1-2cos^2 Y
b) Hence, prove that sin^4 Y - cos^4 = 1-2cos^2 Y
最好詳細列式, thx~~
求兩條數學題答案, 各分a,b part.急!!! thx~
2009-07-30 8:57 pm
回答 (2)
2009-07-30 9:24 pm
✔ 最佳答案
1a) L.H.S.= cos(42-Y)tan(42-Y)= cos (42-Y) * sin (42-Y)/cos(42-Y)
= sin(42-Y)
= cos[90-(42-Y)]
= cos (48 +Y)
Thus, cos (48 +Y)= cos(2Y+18)
48+Y = 2Y+18
Y=30
b) Put y=30
3cosYsin(15+Y)tan 2Y= 3 cos30 sin45 tan 60
= √3/2 * √2/2 * √3
=3√2 / 4
2a) L.H.S.= sin^2 Y - cos^2 Y
=sin^2 Y + cos^2 Y - 2 cos^2 Y
= 1- 2 cos^2 Y
=R.H.S.
therefore, sin^2 Y - cos^2 Y = 1-2cos^2 Y
b) L.H.S.= sin^4 Y - cos^4
= (sin^2 Y)^ 2 -(cos^2 Y)^ 2
= ( sin^2 Y-cos^2 Y)( sin^2 Y+cos^2 Y)
= ( sin^2 Y-cos^2 Y)*1
= 1-2cos^2 Y
=R.H.S.
Therefore, sin^4 Y - cos^4 = 1-2cos^2 Y.
2009-07-30 13:26:03 補充：
" √" 是開方
2009-07-30 13:30:45 補充：
1b)b) Put y=30
3cosYsin(15+Y)tan 2Y
= 3 cos30 sin45 tan 60
= 3*√3/2 * √2/2 * √3
=9√2 / 4
2009-07-30 10:03 pm
1(a)
cos(42-Y)tan(42-Y) = cos(2Y+18),
sin(42-Y) = cos(2Y+18)
cos(90-(42-Y))=cos(2Y+18)
90-42+Y=2Y+18
Y=30
(b) 3cosYsin(15+Y)tan 2Y
=3cos30sin(15+30)tan 60
=3(√3/2)(√2/2)(√3)
=9√2/4
(2)(a)
sin^2 Y - cos^2 Y
=(1-cos)^2 Y - cos^2 Y
=1-2cos^2 Y
(b) sin^4 Y - cos^4 Y
=[(sin^2 Y) - (cos^2 Y)][(sin^2 Y) + (cos^2 Y)]
=[(sin^2 Y) - (cos^2 Y)][1]
=sin^2 Y - cos^2 Y
=1-2cos^2 Y
cos(42-Y)tan(42-Y) = cos(2Y+18),
sin(42-Y) = cos(2Y+18)
cos(90-(42-Y))=cos(2Y+18)
90-42+Y=2Y+18
Y=30
(b) 3cosYsin(15+Y)tan 2Y
=3cos30sin(15+30)tan 60
=3(√3/2)(√2/2)(√3)
=9√2/4
(2)(a)
sin^2 Y - cos^2 Y
=(1-cos)^2 Y - cos^2 Y
=1-2cos^2 Y
(b) sin^4 Y - cos^4 Y
=[(sin^2 Y) - (cos^2 Y)][(sin^2 Y) + (cos^2 Y)]
=[(sin^2 Y) - (cos^2 Y)][1]
=sin^2 Y - cos^2 Y
=1-2cos^2 Y
收錄日期: 2021-04-26 13:44:42
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