數學 複角公式 高手請進

Question

(1)若2COS^2θ -5COSθ +2=0,則COS2θ =???
(2)設tanθ +secθ =3/2 , 且0 < θ < π/2,則下列何者錯誤?
(A)sinθ =5/13 (B)cotθ=5/12 (c)secθ=13/12
(3)若xsinθ=4sinθ/4*cosθ/2 * cosθ/4,則x為什麼=1
ps(4sinθ/4=4*四分之θ 以此類推)

Answer 1

1. 2COS2@ -5COS@ +2=0
(2cos@ - 1)(cos@ - 2) = 0 => cos@ = 1/2 or -2 (不合)
cos(2@) = 2 cos2@ - 1 = 2*(1/2)2 - 1 = -1/2
2. tan@ + sec@ = 3/2
=> sin@/cos@ + 1/cos@ = 3/2
=> sin@ + 1 = (3/2)cos@
=> sin2@ + 2sin@ + 1 = (9/4)cos2@ = (9/4)( 1 - sin2@)
=> 13(sin2@) + 8(sin@) - 5 = 0
=> (13 sin@ - 5)(sin@ + 1) = 0 => sin @ = 5/13 or -1 (不合)
=> (A)sin@=5/13 yes. (B)cot@=5/12 no. (c)sec@=13/12 yes.
3. xsin@
= x [2sin(@/2)*cos(@/2)] (二倍角)
= x {2[2sin(@/4)*cos(@/4)] * cos(@/2)}
= 4x * sin(@/4)*cos(@/2) * cos(@/4)
= 4sin(@/4)*cos(@/2) * cos(@/4)
=> 4x = 4 => x = 1
如果有問題, 請來函討論. 不然, 我可能會錯失你再補充的疑點.