A maths

Note that the answer will be in radian

Ans 1. Let α be the subsidiary angle.

In △ABC

cos α=1 √2 ⊿1
sin α=1 1

r = √2 α = π/4

√2 (cosxcos α + sinxsin α ) = √2
√2 cos(x- π/4 ) = √2
cos (x- π/4) = 1
x- π/4 = 2nπ
∴ x = 2nπ + π/4


2. 4sinx+5cosx=3

Let β be the subsidiary angle.

In △ABC
cos β= 5 √41 ⊿ 4
sin β= 4 5

r = √41 β = 0.6747 (corr. to 4 sig. fig. )

5(cosxcos β) +4(sinxsinβ ) = 3
√41cos(x-0.6747) = 3
cos(x-0.6747) = 3/ √41
x-0.6747 = 2nπ±1.083 (corr.to 4 sig.fig.)

x = 2nπ+1.083 +0.6747 or x =2nπ+0.6747-1.083
∴ x = 2nπ + 1.76 (corr. to 3 sig. fig.) or x= 2nπ - 0.408(corr. to 3sig. fig)


3. tanx = cotx
tanx = tan (π/2 - x )
x = nπ + π/2 - x
∴ 2x = nπ + π/2
x = nπ/2 + π/4

2007-04-21 22:12:28 補充：
Sorry~ the answer is not clear enough1.cos α=1 sin α=1 √2 ⊿1 12.cos β= 5 sin β= 4 √41 ⊿ 4 5

2007-04-21 22:15:13 補充：
下面那位答題者√(1 1)sin(x π/4)=√2sin(x π/4)=0　　← (應該是1吧?)

* ^ is the index no. and (+-) is positive or negative
1. √2(1/√2sinx+1/√2cosx)=√2
(sin45sinx+cos45cosx)=1
(cos45cosx+sin45sinx)=1
cos(x-45)=1
x-45=360n (where n is a natural no.)
x=360n+45
2. √41(4/√41sinx+5√41cosx)=3
√41(cos51.34sinx+sin51.34cosx)=3
√41(sin(51.34+x)=3
sin(51.34+x)=3/√41
51.34+x=180n+(-1)^n(27.94) (where n is a natural no.)
x=180n+(-1)^n(-23.4)
x=180n+(-1)^n+1(23.4)
3. tanx=1/tanx
tan^2x=1
tan^2x-1=0
tanx=1 or -1
tanx=180n(+-)45 (where n is a natural no.)