高考数学一轮复习同角三角函数基本关系式与诱导公式练习含答案 下载本文

解析 ∵sin(π+θ)=-3cos(2π-θ), ∴-sin θ=-3cos θ, ππ

∴tan θ=3,∵|θ|<2,∴θ=3. 答案 D

14.若sin θ,cos θ是方程4x2+2mx+m=0的两根,则m的值为( ) A.1+5 C.1±5

B.1-5 D.-1-5

mm

解析 由题意知sin θ+cos θ=-2,sin θ·cos θ=4. 又sin θ+cos θ()=1+2sin θcos θ,

2

m2m

∴4=1+2,解得m=1±5. 又Δ=4m2-16m≥0, ∴m≤0或m≥4,∴m=1-5. 答案 B

15.sin21°+sin22°+…+sin290°=________.

解析 sin21°+sin22°+…+sin290°=sin21°+sin22°+…+sin244°+sin245°+cos244°+cos243°+…+cos21°+sin290°=(sin21°+cos21°)+(sin22°+cos22°)+…191

+(sin244°+cos244°)+sin245°+sin290°=44+2+1=2. 91答案 2 ?π??5π??2π?

?+sin??=________. 16.已知cos?-θ?=a,则cos?+θ-θ

?6??6??3??5π???π???π?

解析 ∵cos?+θ?=cos?π-?-θ??=-cos?-θ?=-a.

?6???6???6??π?π???2π?π??=cos???=a, sin?-θ?=sin?+?-θ??-θ?2?3??6??6???

?5π??2π?

∴cos?+θ?+sin?-θ?=0.

?6??3?答案 0