Prove the following identities.
1. tan(90-θ) / 1+tan^2(90-θ) = sinθcosθ
2. 1-sin^2 θ / 1- cosθ = (1+cosθ) tan^2(90-θ)
Find the value of the following expression.
1. Given that tan(90-θ)=1/2, find the value of sin(90-θ)/1-tanθ + 2cos(90-θ)
F.3 Trigonometric Relations
2011-05-06 7:05 am
回答 (2)
2011-05-06 7:40 am
2011-05-06 7:26 am
1.tan(90-θ)/(1+tan^2(90-θ))
=cotθ/(1+cot^2θ)
=(cosθ/sinθ)/(1+(cosθ/sinθ)^2)
=(cosθsinθ)/(sin^2θ+cos^2θ)
=sinθcosθ
2.1-sin^2θ/(1-cosθ)
=(1-sin^2θ)(1+cosθ)/(1-cos^2θ)
=(1-sin^2θ)(1+cos)/sin^2θ
=(1+cosθ)(cosθ/sinθ)^2
=(1+cosθ)(cotθ)^2
=(1+cosθ)tan^2(90-θ)
3.tan(90-θ)=1/2--> cotθ=1/2, tanθ=2
1).sin(90-θ)=-1/√5,cos(90-θ)=-2/√5
2).sin(90-θ)=1/√5,cos(90-θ)=2/√5
sin(90-θ)/(1-tanθ)+2cos(90-θ)
=-1/√5/(-1)+2(-2/√5)
=-3/√5
或=1/√5/(-1)+2(2/√5)
=3/√5
2011-05-06 01:02:47 補充：
注意θ在此處沒有限定
所以sin,cos有可能皆是負的
=cotθ/(1+cot^2θ)
=(cosθ/sinθ)/(1+(cosθ/sinθ)^2)
=(cosθsinθ)/(sin^2θ+cos^2θ)
=sinθcosθ
2.1-sin^2θ/(1-cosθ)
=(1-sin^2θ)(1+cosθ)/(1-cos^2θ)
=(1-sin^2θ)(1+cos)/sin^2θ
=(1+cosθ)(cosθ/sinθ)^2
=(1+cosθ)(cotθ)^2
=(1+cosθ)tan^2(90-θ)
3.tan(90-θ)=1/2--> cotθ=1/2, tanθ=2
1).sin(90-θ)=-1/√5,cos(90-θ)=-2/√5
2).sin(90-θ)=1/√5,cos(90-θ)=2/√5
sin(90-θ)/(1-tanθ)+2cos(90-θ)
=-1/√5/(-1)+2(-2/√5)
=-3/√5
或=1/√5/(-1)+2(2/√5)
=3/√5
2011-05-06 01:02:47 補充：
注意θ在此處沒有限定
所以sin,cos有可能皆是負的
收錄日期: 2021-04-23 20:43:43
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