1、已知cos(-100°)=k,则tan80°=?

1个回答

  • 1、cos(80°)=-cos(80°-180°)=-k,sin(80°)=√{1-k^2)}(舍负),tan80°=sin80°/cos80°=-√{1-k^2}/k

    2、α∈(π/2+2kπ,π+2kπ),则α/2∈(π/4+kπ,π/2+kπ)=(π/4+2kπ,π/2+2kπ)∪(5π/4+2kπ,3π/2+2kπ),又因为其余弦值为负,故α/2属于第3象限

    3、由于cost+cos(180°-t)=0,cos90°=0,cos180°=-1故原式值为-1(类似等差数列求和的思想,前后两两配对).事实上,若最后一项不为cos180°,仍然可以求和.利用积化和差公式,先乘以sinα,下面求α.由于sinαcosβ=[sin(β+α)-sin(β-α)]/2,要使sinαcosβ与sinαcos(β+1°)求和抵消掉一项,则β-α=β+1°+α,故α=-0.5°.所以原式=[sin(1°-0.5°)-sin(180°+0.5°)]/[2*sin(-0.5°)]=-1

    4、由于sin(3π+β)=0(lg1=0),则sinβ=0,cosβ=1或-1({cos(π-β)-1}如果为分母,则cosβ=1,否则分母为0),两种情况分别代入计算可得结果:若{cos(π-β)-1}为分母,则原式=0.5;否则,原式=1+cosβ=2或0