Maths Problem

tan(90-x)sinx+cosx=√2

[sin(90-x)/cos(90-x)]sinx + cosx = √2

[cosx/sinx]sinx + cosx = √2

cosx + cosx = √2

cosx = 1/√2

x = 45⁰

2013-12-21 15:21:10 補充：
[(tanx+sinx)^2-(tanx-sinx)^2]^2=16(tanx+sinx)(tanx-sinx)

L.H.S. = [(tanx+sinx)^2-(tanx-sinx)^2]^2

Using a² - b² = (a+b)(a-b)

[(tanx+sinx)²-(tanx-sinx)²]² = {[(tanx+sinx)+(tanx-sinx)][(tanx+sinx)-(tanx-sinx)]}²

={[2tanx][2sinx]}²

=16tan²xsin²x

2013-12-21 15:26:59 補充：
R.H.S. = 16(tanx+sinx)(tanx-sinx)

=16(sinx/cosx+sinx)(sinx/cosx-sinx)

=16[(sinx+sinxcosx)/cosx][sinx-sinxcosx)/cosx]

=16sinx[(1+cosx)/cosx]sinx[(1-cosx)/cosx]

=16(sin²x/cos²x)(1+cosx)(1-cosx)

=16tan²x(1-cos²x)

=16tan²xsin²x

=L.H.S.

2013-12-21 16:12:15 補充：
3.a)
2/DC = sin45⁰ = 1/√2
DC = 2√2

b)
AC = AE + EC
AE/DE = tan60⁰ = √3
AE = (√3)(DE) = 2√3

EC/DE = tan45⁰ = 1
EC = DE = 2

AC = AE + EC = 2√3 + 2

c)
AB/AC = cos45⁰ = 1/√2
AB = (1/√2)(AC)
=(1/√2)(2√3 + 2)
=(√2)(√3 + 1)

d)


2013-12-21 16:19:07 補充：
Before anything, since AB = BC and angle ABC = 90⁰, angle ACB = angle BAC = 45⁰.

2013-12-21 16:26:36 補充：
d)

Draw a line from A to somewhere F and such that AF // BC
Draw a line from D to meet AF at G such that DG // BA

angle GAD = 75⁰
DG = AB = (√2)(√3 + 1)

2/AD = cos60⁰ = 1/2
AD = 4

sin75⁰ = DG/AD = [(√2)(√3 + 1)]/4

= (1+√3)/(2√2)