Question

FH

求16第二小題詳解！急

16. 如右圖, AB 為直徑,且 AB =2,若PA-PB=-1,∠BQP=0,
則:
(1) sincos 日之值=
(2) sin°8+cos°0之值=
64
sin =
再 cosa=
COSO-
sine =
(sin²est (cos³0)³
= (sin ²0+ cos²o) - 3 sin ²0 costa
(sin²e + cos²0) = ³²³-3x²² 3³ x
= 1-
一
PA - PB =
Oric (coso - sin o) ² = =
A
cos²a - 2205g Siha + sine - /
1-2 cosasina = 1/2
acasa sina = 3
10.
|B

Answer 1

K

看圖

FH

請問減3xsincos怎麼來的？

K

圖