Given that cosx-√3sinx=rcos(x+α),where r>0 and α is acute ,find the value of r² and tanα.
Hence solve the equation cosx-√3sinx=1 for -180° <180° .
Write down the least valur of (1/(cosx-√3sinx)²)
Plx show your steps clearly as soon as posible
ans:r²=4,tanα=√3
0°,-120°
1/4
(2) Trigonometric functions
2007-03-02 3:29 am
回答 (2)
2007-03-02 6:44 am
✔ 最佳答案
Please find the answers as follows:圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Maths/Crazytrigo46.jpg
參考: My Maths knowledge
2007-03-02 6:50 am
cosx-√3sinx=rcos(x+α)
=rcosxcosα-rsinxsinα
=(rcosα)cosx-(rsinα)sinx
comparing the coefficient,
rcosα=1 and rsinα=√3
r² =1²+√3²
r²=4
rsinα/rcosα=√3/1
tanα=√3
thus,r²=4 and tanα=√3
Hence solve the equation cosx-√3sinx=1 for -180° <180° .
cosx-√3sinx=1
2cos(x+60)=1
cos(x+60)=1/2
x+60=60 or 300
x=0 or 240
x=0 or -120
Write down the least valur of (1/(cosx-√3sinx)²)
1/(cosx-√3sinx)²
={1/[2cos(x+60)]²}
=1/4(1)
=1/4
=rcosxcosα-rsinxsinα
=(rcosα)cosx-(rsinα)sinx
comparing the coefficient,
rcosα=1 and rsinα=√3
r² =1²+√3²
r²=4
rsinα/rcosα=√3/1
tanα=√3
thus,r²=4 and tanα=√3
Hence solve the equation cosx-√3sinx=1 for -180° <180° .
cosx-√3sinx=1
2cos(x+60)=1
cos(x+60)=1/2
x+60=60 or 300
x=0 or 240
x=0 or -120
Write down the least valur of (1/(cosx-√3sinx)²)
1/(cosx-√3sinx)²
={1/[2cos(x+60)]²}
=1/4(1)
=1/4
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