cos²80°+sin²50°-sin190°·cos320°

cos²80°+sin²50°-sin190°·cos320°
cos²80°+sin²50°-sin190°·cos320°

cos²80°+sin²50°-sin190°·cos320°
= (sin10)^2 + (sin50)^2 + sin10sin50
= (sin10)^2 + (sin50)^2 - 2sin10sin50cos120
想像一个三角形,三个角分别是10,50,120,
对应的三边长sin10,sin50,sin120
由余弦定理得
(sin120)^2 = (sin10)^2 + (sin50)^2 - 2sin10sin50cos120
所以原式就等于 (sin120)^2 = 3/4

cos^2 80°+sin^2 50°-sin190°cos320
=sin10^2+sin50^2-sin(180+10)cos(360-40)
=sin10^2+cos40^2+sin10cos40
设sin10=x,cos40=y
=x^2+y^2+xy
=(x+y)^2-xy