Prove the following trigonometric identities
cos^2θ - sin^2θ / 1+ 2sinθcosθ = 1-tanθ / 1+ tanθ
F.3 Trigonometric identites
2011-03-19 6:55 am
回答 (2)
2011-03-19 8:05 am
✔ 最佳答案
cos^2θ - sin^2θ / 1+ 2sinθcosθ = 1-tanθ / 1+ tanθ ___________________
RHS:
1-tanθ / 1+ tanθ
(cosθ-sinθ/cosθ)/(cosθ+sinθ/cosθ)
(cosθ-sinθ/cosθ)(cosθ/sinθ+cosθ)
(cosθ-sinθ/sinθ+cosθ)
(1/cosθ)(cosθ-sinθ)/(1/cosθ)(sinθ+cosθ)
1-tanθ/1+tanθ
=LHS
2011-04-02 12:03:38 補充：
cos^2θ - sin^2θ / 1+ 2sinθcosθ = 1-tanθ / 1+ tanθ
LHS:
cos^2θ - sin^2θ / 1+ 2sinθcosθ
(cosθ-sinθ )/(sinθ +cosθ )
______________________
RHS:
1-tanθ / 1+ tanθ
(cosθ-sinθ/cosθ)/(cosθ+sinθ/cosθ)
(cosθ-sinθ/cosθ)(cosθ/sinθ+cosθ)
(cosθ-sinθ/sinθ+cosθ)
2011-03-19 8:14 am
LHS= (cos²θ-sin²θ) / ( 1+ 2sinθcosθ)
= (cosθ- sinθ)(cosθ+ sinθ) / ( cos²θ+ 2sinθcosθ+ sin²θ)
= (cosθ- sinθ)(cosθ+ sinθ) / (cosθ+ sinθ)²
= (cosθ- sinθ )/ (cosθ+ sinθ)
= [(cosθ- sinθ)/ cosθ] / [(cosθ+ sinθ)/ cosθ]
= (1-tanθ) / (1+ tanθ)
= RHS
Formula being used :
A²-B² = (A+B)(A-B)
(A+B)²= A²+2AB+B²
sin²A+ cos²A = 1
= (cosθ- sinθ)(cosθ+ sinθ) / ( cos²θ+ 2sinθcosθ+ sin²θ)
= (cosθ- sinθ)(cosθ+ sinθ) / (cosθ+ sinθ)²
= (cosθ- sinθ )/ (cosθ+ sinθ)
= [(cosθ- sinθ)/ cosθ] / [(cosθ+ sinθ)/ cosθ]
= (1-tanθ) / (1+ tanθ)
= RHS
Formula being used :
A²-B² = (A+B)(A-B)
(A+B)²= A²+2AB+B²
sin²A+ cos²A = 1
收錄日期: 2021-04-13 17:52:20
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