Question

急求数学必修四课后P143习题3.2答案，在线等！！ A组 1.求证 (2)tan a/2 -1/（tan a/）= -2/tan a

（3)tan (x/2+π/4)+tan(x/2-π/4)=2tanx
B组
1.求证
（1）3+cos4a-4cos2a=8sin^4 a (sin的右上方还有个4次方）
（2）tanatan2a/(tan2a-tana)+根号3（sin^2 a-cos^2 a)=2sin(2a-π/3)

Answer 1

（1）3+cos4a-4cos2a=3+2cos2a^2-1-4cos2a=2(1+cos2a^2-2cos2a)=2(1-cos2a)^2=2(2sina^2)^2=8sina^4
(3) tan(x/2+π/4)+tan(x/2-π/4)
=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]
=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]
=[(tan(x/2)+1)^2-(tan(x/2)-1)^2]/[1-(tan(x/2))^2]
=4tan(x/2)/[1-(tan(x/2))^2]
=4tan(x/2)/[1-(tan(x/2))^2]=2{tan(x/2)+tan(x/2)/[1-(tan(x/2))×(tan(x/2))]}
=2tanx

Answer 2

2）做长方形ABCD （AD＞AB）在BC上取一点P使∠APB＝a ∠PDC＝a/2
则tan a＝AB/BP tan a/2＝PC/CD＝PC/AB AP＝AD
∴AB^2+BP^2＝AP^2＝AD^2＝（BP+PC）^2
∴AB^2＝2BP×PC+PC^2 同除以（AP×PC）
得 AB/PC＝2BP/AB+PC/AB
把tan a＝AB/BP tan a/2＝PC/CD＝PC/AB 代入
1/（tan a/2）＝2/（tan a）+ tan a/2
即tan a/2 - 1/（tan a/2）＝-2/（tan a）

Answer 3

（3)tan (x/2+π/4)+tan(x/2-π/4)=2tanx
=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]
=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]
=[(tan(x/2)+1)^2-(tan(x/2)-1)^2]/[1-(tan(x/2)^2]
=4tan(x/2)/[1-(tan(x/2)^2]
=4tan(x/2)/[1-(tan(x/2))^2]=2[2tan(x/2)/1-tan(x/2)^2]=2tanx

Answer 4

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