A.Maths 小問題之 trigonometry

a) Using the formula sin 2Φ = 2 sinΦcosΦ , find the range of the values of sinΦcosΦ

The range of sin 2Φ is -1 to 1 So the range of the values of sinΦcosΦ is -1/2 to 1/2

-1/2＜=sinΦcosΦ＜=1/2

b) Given sinΦ+cosΦ = sinΦcosΦ , find the value of sinΦcosΦ

sinΦ+cosΦ = sinΦcosΦ
(sinΦ+cosΦ)^2 = (sinΦcosΦ)^2
1+2sinΦcosΦ = (sinΦcosΦ)^2
Let sinΦcosΦ=x, then x^2-2x-1=0
x=1/2(2-√8)=1-√2
sinΦcosΦ=1-√2

(c)
sinΦ+cosΦ = sinΦcos^2Φ
sinΦ+cosΦ = (1-√2)cosΦ
tanΦ+1=1-√2
tanΦ=-√2
Φ=180-tan^(-1)√2 or 360-tan^(-1)√2

2008-03-17 10:37:24 補充：
呀﹐我(c)part個答案代回不正確。不過既然用hence﹐應該要用到(b)﹐飛天個答案正案﹐但是無用(b)﹐而且您人手如何解三次方程。所以請容我慢慢想一想再給您一個答覆

part (b) sinΦ+cosΦ = sinΦcosΦ
It's not true. You can take Φ=45 and get it.
Any typing mistakes here?