三角函數證明y=-t^2/2 + 1/2

只需用簡單的三角函數恆等式 tanθ = sinθ/cosθ 和四則混算便可。

左式
= y
= [(cos²θ - sin²θ) / 2sinθcosθ ] x tanθ
= (1/2)[(cos²θ/sinθcosθ) - (sin²θ/sinθcosθ)] x tanθ
= (1/2)[(cosθ/sinθ) - (sinθ/cosθ)] x tanθ
= (1/2)[(1/tanθ) - tanθ] x tanθ
= (1/2)[(1/tanθ) - (tan²θ/tanθ)] x tanθ
= (1/2)[(1 - tan²θ) / tanθ] x tanθ
= (1/2)(1 - tan²θ)
= (-1/2)(tan²θ - 1)
= (-1/2)(t² - 1)
= -t²/2 + 1/2
= 右式

cos(2θ) = cos²θ-sin²θ
sin(2θ) = 2sinθcosθ
tan(2θ) = [2tanθ] / [1-tan²θ]

y = [(cos²θ-sin²θ)/2sinθcosθ ] * tanθ

y = [cos(2θ)/sin(2θ)] * tanθ

y = [1/tan(2θ)]* tanθ

y = [(1-tan²θ)/(2tanθ)]* tanθ

y = [(1-tan²θ)/2]

y = 1/2 - tan²θ/2

y = 1/2 - t²/2

y = - t²/2 + 1/2