2018-2019学年北师大版必修四 二倍角的三角函数 课时作业
2018-2019学年北师大版必修四     二倍角的三角函数  课时作业第1页

§3 二倍角的三角函数

A组 基础巩固

1.若tan α=3,则sin2α/(cos^2 α)的值等于(  )

A.2 B.3 C.4 D.6

解析sin2α/(cos^2 α)=2sinαcosα/(cos^2 α)=2sinα/cosα=2tan α=6.

答案D

2.√(1+cos100"°" )-√(1"-" cos100"°" )等于(  )

A.-2cos 5° B.2cos 5°

C.-2sin 5° D.2sin 5°

解析原式=√(2cos^2 50"°" )-√(2sin^2 50"°" )

  =√2(cos 50°-sin 50°)

  =2(√2/2 cos50"°-" √2/2 sin50"°" )

  =2sin(45°-50°)

  =-2sin 5°.

答案C

3.cosπ/17·cos2π/17·cos4π/17·cos8π/17的值为(  )

A.1/2 B.1/4 C.1/8 D.1/16

解析乘以(sin π/17)/(sin π/17),利用倍角公式化简得1/16.

答案D

4.已知3/2π<α<2π,化简√(1/2+1/2 √(1/2+1/2 cos2α)) 的结果为(  )

A.sinα/2 B.-sinα/2

C.cosα/2 D.-cosα/2

解析∵3π/2<α<2π,∴3π/4<α/2<π,

  ∴cos α>0,cosα/2<0,

  ∴原式=√(1/2+1/2 √((1+cos2α)/2)) √(1/2+1/2 √(cos^2 α))

  =√((1+cosα)/2)=√(cos^2 α/2)=-cosα/2.

答案D

5.若sin 2α=24/25,0<α<π/2,则√2cos(π/4 "-" α)的值为(  )

A.1/5 B.-1/5 C.±1/5 D.7/5

解析(sin α+cos α)2=1+sin 2α=49/25,

  因为0<α<π/2,所以sin α+cos α=7/5,

  则√2cos(π/4 "-" α)=√2×√2/2(cos α+sin α)=7/5.

答案D

6.若(sinα+cosα)/(sinα"-" cosα)=1/2,则tan 2α=(  )

A.-3/4 B.3/4 C.-4/3 D.4/3