I have 2 math questions... plz help !!!!?

Question

1. for 0 < B < 2pi, determine all solutions of SinB + cosB = SinBcosB

2. Simplify the function f(x) = sin3x cscx - cos 3x secx

plz help me out....

thx

Answer 1

Use trigonometric identities
There's a whole bunch listed at:
http://en.wikipedia.org/wiki/Trigonometric_identities

For example, rewrite number 2 as:

Sin(3x)/Sin(x) - Cos(3x)/Cos(x)

Sin(3x) = Sin(2x+x) = Sin(2x)Cos(x) + Cos(2x)Sin(x)

and

Cos(3x) = Cos(2x+x) = Cos(2x)Cos(x) - Sin(2x)Sin(x)
Then you use:

Sin(2x) = 2Sin(x)Cos(x)
and
Cos(2x) = Cos^2(x) - Sin^2(x)

so that
Sin(3x) = 2Sin(x)Cos(x)Cos(x) + [Cos^2(x) - Sin^2(x)]Sin(x)
and
Cos(3x) = [Cos^2(x) - Sin^2(x)]Cos(x) - 2Sin(x)Cos(x)Sin(x)

---

f(x) = Sin(3x)/Sin(x) - Cos(3x)/Cos(x) =
{2Sin(x)Cos(x)Cos(x) + [Cos^2(x) - Sin^2(x)]Sin(x)}/Sin(x) - {[Cos^2(x) - Sin^2(x)]Cos(x) - 2Sin(x)Cos(x)Sin(x)}/Cos(x)
=
{ 3Cos^2(x) - Sin^2(x) } - { Cos^2(x) - 3Sin^2(x) }
=
2 Cos^2(x) + 2 Sin^2(x)
= 2 {Cos^2(x) + Sin^2(x)}
=
2 {1} = 2

Answer 2

no clue not very good with math