F.4 Additional Mathematics

2009-04-11 7:23 am

Answer only ~~~ RED With Steps~~~BLUE
2. sin( q – p/2) = cos p/11, 0 q 2p
4. tan( q – p/7 ) = cot 15p/7, 0 q 2p
5. 2sin 2q = (root 3)sin 2q cos q, –p q p

8. 2 sin q cos q – sin q + 4 cos q + 2 = 0, –2p q 0
14. tan 1/2 q – 2 cot 1/2 q + 1 = 0, 0 q 2p
Thank you!

更新1:

Dear eelyw, what do u mean p and pi? i don't understand.

回答 (1)

用戶10012

2009-04-11 3:23 pm

✔ 最佳答案

Q2 and Q4. What is p? Could it be pi?
Q5.
sin 2q(2 - sqrt3 cos q) = 0
that is sin 2q = 0 .................(1) and
2 - sqrt3 cos q = 0..........(2)
From (1) q = 0 degree ( or use the general solution formula)
From (2) cos q = 2/sqrt3 ( rej).
Q8.
Could it be -2, not +2?
Q14.
let tan (q/2) = x, so
x - 2/x + 1 = 0
x^2 + x - 2 = 0
(x - 1)(x + 2) = 0
that is tan (q/2) = 1............(1) and
tan(q/2) = -2...............(2)
From (1) q/2 = pi/4, q = pi/2.
From (2) q/2 = arc tan(-2), q = 2 arctan (-2).

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