1Evaluate:
A,tan^2 30°·sin^2 45°/cos^2 60°
B,sin^2 45°-cos^2 45°+sin30°·cos60°+tan60°
2Prove:
A,√(1+tan^2θ)=1/cosθ
B,5+4sin(90°-θ)-sin^2 θ=(cosθ+2)^2
數條英數題(中二級)
2009-07-11 3:18 am
回答 (2)
2009-07-11 4:00 am
✔ 最佳答案
1Atan230sin245/cos260
=(1/√3)2(1/√2)2/(1/2)2
=(1/3)(1/2)/(1/4)
=2/3
1B
sin245- cos245+sin30cos60 + tan60
=(1/√2)2 - (1/√2)2 +(1/2)(1/2) + √3
=1/4 + √3
2A
√(1+tan2θ)
=√(1+sin2θ/cos2θ)
=√[(cos2θ+sin2θ)/cos2θ]
=√(1/cos2θ)
=1/cosθ
2B
5+4sin(90-θ)-sin2θ
= 5 + 4cosθ - (1 - cos2θ)
= 4 + 4cosθ + cos2θ
=(cosθ+2)2
2009-07-11 4:15 am
1A. tan^2 30°·sin^2 45°/cos^2 60°
= (tan 30°)^2 ( sin45°)^2 / ( cos60°)^2
= ( 1/ 「3 ) ^2 ( 1/ 「2) ^2 / ( 1/2 )^2
= 1/3 ( 1/2 ) / ( 1/4 )
= 2/3
1B. sin^2 45°-cos^2 45°+sin30°·cos60°+tan60
= ( 1/ 「2) ^2 - ( 1/ 「2) ^2 + ( 1/2 )(1/2 ) + 「3
= 1/4 + 「3
= ( 1+4「3 ) /4
2A. LHS = √(1+tan^2θ)
= √(1+ ( sin^2θ / cos^2θ )
= √ ( ( cos^2θ +sin^2θ ) / cos^2θ )
= √( 1/ cos^2θ )
= 1/cosθ = RHS
2B. LHS = 5+4sin(90°-θ)-sin^2 θ
= 5+ 4cosθ - sin^2θ
= 1-sin^2θ + 4 + 4cosθ
= cos^2θ+ 4cosθ +4
= ( cosθ + 2 ) ^2 = RHS
= (tan 30°)^2 ( sin45°)^2 / ( cos60°)^2
= ( 1/ 「3 ) ^2 ( 1/ 「2) ^2 / ( 1/2 )^2
= 1/3 ( 1/2 ) / ( 1/4 )
= 2/3
1B. sin^2 45°-cos^2 45°+sin30°·cos60°+tan60
= ( 1/ 「2) ^2 - ( 1/ 「2) ^2 + ( 1/2 )(1/2 ) + 「3
= 1/4 + 「3
= ( 1+4「3 ) /4
2A. LHS = √(1+tan^2θ)
= √(1+ ( sin^2θ / cos^2θ )
= √ ( ( cos^2θ +sin^2θ ) / cos^2θ )
= √( 1/ cos^2θ )
= 1/cosθ = RHS
2B. LHS = 5+4sin(90°-θ)-sin^2 θ
= 5+ 4cosθ - sin^2θ
= 1-sin^2θ + 4 + 4cosθ
= cos^2θ+ 4cosθ +4
= ( cosθ + 2 ) ^2 = RHS
收錄日期: 2021-04-23 23:22:55
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https://hk.answers.yahoo.com/question/index?qid=20090710000051KK01743