Prove the following identity
sin^4θ - cos^4θ 恆等於 2sin^2θ -1
Maths..
2010-09-12 1:24 pm
回答 (4)
2010-09-12 4:45 pm
✔ 最佳答案
(sinθ)^4-(cosθ)^4=[(sinθ)^2+(cosθ)^2][(sinθ)^2-(cosθ)^2]
=(1)[(sinθ)^2-(cosθ)^2] (since [(sinθ)^2+(cosθ)^2]=1)
=(sinθ)^2-[1-(sinθ)^2]
=2(sinθ)^2-1
2010-09-12 8:56 pm
sin^4θ - cos^4θ 恆等於 2sin^2θ -1
_________________________
LHS
sin^4θ - cos^4θ
(sin^2X-cos^2X)(sin^2X+cos^2X)
(sin^2X-cos^2X)
_____________
RHS
2sin^2θ -1
2sin^2X-cos^2X-sin^2X
sin^2X-cos^2X
LHS
_________________
sin^2X+cos^2X=1
_________________________
LHS
sin^4θ - cos^4θ
(sin^2X-cos^2X)(sin^2X+cos^2X)
(sin^2X-cos^2X)
_____________
RHS
2sin^2θ -1
2sin^2X-cos^2X-sin^2X
sin^2X-cos^2X
LHS
_________________
sin^2X+cos^2X=1
2010-09-12 7:24 pm
You should remember:
sin^2 θ + cos^2 θ = 1
Hence, (sin^2θ+cos^2θ)(sin^2θ-cos^2θ)
= 1(sin^2θ-cos^2θ)
= (sin^2θ-cos^2θ)
sin^2 θ + cos^2 θ = 1
Hence, (sin^2θ+cos^2θ)(sin^2θ-cos^2θ)
= 1(sin^2θ-cos^2θ)
= (sin^2θ-cos^2θ)
2010-09-12 3:13 pm
sin^4θ-cos^4θ
=(sin^2θ+cos^2θ)(sin^2θ-cos^2θ)
=sin^2θ-cos^2θ
=sin^2θ-(1-sin^2θ)
=sin^2θ-1+sin^2θ
=2sin^2θ-1
=(sin^2θ+cos^2θ)(sin^2θ-cos^2θ)
=sin^2θ-cos^2θ
=sin^2θ-(1-sin^2θ)
=sin^2θ-1+sin^2θ
=2sin^2θ-1
參考: 我
收錄日期: 2021-04-13 17:31:44
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