已知角a的终边在射线y=-2x(x≥0)上,则sin（2a+5π/4）的值是_____
问题描述：
已知角a的终边在射线y=-2x(x≥0)上,则sin（2a+5π/4）的值是_____
答
解法一：
sina=y/r=-2x/r
cosa=x/r
r^2=x^2+y^2=x^2+4x^2=5x^2
sin(2a+5π/4)
=sin2acos5π/4+cos2asin5π/4
=2sinacosacos5π/4+(2cos^2a-1)sin5π/4
=-√2sinacosa-√2/2(2cos^2a-1)
=-√2sinacosa-√2cos^2a+√2/2
=-√2(-2x/r)(x/r)-√2(x/r)^2+√2/2
=2√2x^2/r^2-√2(x/r)^2+√2/2
=√2(2x^2/r^2-x^2/r^2)+√2/2
=√2x^2/r^2+√2/2
=√2x^2/(5x^2)+√2/2
=√2/5+√2/2
=7√2/10
解法二：
tana=y/x=-2x/x=-2
sin(2α+5π/4)
=sin2αcos5π/4+cos2αsin5π/4
=-√2/2(sin2a+cos2a)
=-√2/2[2tana/(1+tan^2a)+(1-tan^2a)/(1+tan^2a)]
=-√2/2(2tana+1-tan^2a)/(1+tan^2a)
=-√2/2(-4+1-4)/(1+4)
=7√2/10