Question

根号下[sin^2(500°)+sin^2(770°)-cos^2(1620°-x)]，(180°<x<270°)

很急！

Answer 1

(180°<x<270°)
sinx<0

√[sin^2(500°)+sin^2(770°)-cos^2(1620°-x)]
=√[sin^2(360°+140°)+sin^2(720°+50°)-cos^2(1440°+180°-x)]
=√[sin^2(140°(+sin^2(50°)-cos^2(180°-x)]
=√[sin^2(180°-40°)+sin^2(50°)-cos^2(180°-x)]
=√[sin^2(40°)+sin^2(90°-40°)-cos^2(180°-x)]
=√[sin^2(40°)+cos^2(40°)-cos^2(180°-x)]
=√[1-(-cosx)^2]
=√[1-(cosx)^2]
=√[(sinx)^2]
=-sinx

Answer 2

500=360+140
140=180-40
770=720+50
1620=1800-180
(sin500)^2+(sin770)^2-(cos(1620-x))^2
=(sin40)^2+(sin50)^2-(cosx)^2
=1-(cosx)^2
=(sinx)^2
√(sinx)^2=|sinx|=-sinx