Find the value of it.
sec[(5π)/4+θ]
Trigonometric funtion~快來
2007-01-22 7:54 am
回答 (2)
2007-01-22 9:18 pm
sec[(5π)/4 + θ]
= 1/ cos[(5π)/4 + θ]
= -1/ cos[π/4 + θ]
= -1/ [cos(π/4)cosθ-sin(π/4)sinθ]
= -root2/ (cosθ-sinθ)
= 1/ cos[(5π)/4 + θ]
= -1/ cos[π/4 + θ]
= -1/ [cos(π/4)cosθ-sin(π/4)sinθ]
= -root2/ (cosθ-sinθ)
2007-01-22 5:58 pm
sec[(5π)/4+θ]
=sec[π + (π/4 + θ)]
= -sec(π/4 + θ)
= -1/ cos(π/4 + θ) ............(1)
cos(π/4 + θ)
= cosπ/4cosθ - sinπ/4sinθ
= (√2/2) cosθ - (√2/2) sinθ
= √2/2 [cosθ - sinθ]...........(2)
Put (2) into (1), we get
sec[(5π)/4+θ]
= -1/ [√2/2 (cosθ - sinθ)]
= -2/ [√2 (cosθ - sinθ)]
= -√2/ (cosθ - sinθ)
= √2/ (sinθ - cosθ)
=sec[π + (π/4 + θ)]
= -sec(π/4 + θ)
= -1/ cos(π/4 + θ) ............(1)
cos(π/4 + θ)
= cosπ/4cosθ - sinπ/4sinθ
= (√2/2) cosθ - (√2/2) sinθ
= √2/2 [cosθ - sinθ]...........(2)
Put (2) into (1), we get
sec[(5π)/4+θ]
= -1/ [√2/2 (cosθ - sinθ)]
= -2/ [√2 (cosθ - sinθ)]
= -√2/ (cosθ - sinθ)
= √2/ (sinθ - cosθ)
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070121000051KK05605