Maths Problem_10－oob73zj92b的部落格

標題:

Maths Problem

發問:

1)Find the value of x if tan(90-x)sinx+cosx=√22)Prove that [tanx+sinx)^2-(tanx-sinx)^2]^2=16(tanx+sinx)(tanx-sinx)3)In triangleABC, AB=CB and angleABC =90degree . D and E are points on BC and AC respectively such that DE is perpendicular AC . It is given that angleADE=60 degree and DE=2cm.(a)Find... 顯示更多 1)Find the value of x if tan(90-x)sinx+cosx=√2 2)Prove that [tanx+sinx)^2-(tanx-sinx)^2]^2=16(tanx+sinx)(tanx-sinx) 3)In triangleABC, AB=CB and angleABC =90degree . D and E are points on BC and AC respectively such that DE is perpendicular AC . It is given that angleADE=60 degree and DE=2cm. (a)Find DC (b)Show that AC =(2+2√3) (c)Using the result of (b), find AB (d)Using the result of (c),or otherwise , show that sin75=(1+√3)/(2√2) NEED STEP,PLZ!!!

最佳解答:

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 = 45? 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 = sin45? = 1/√2 DC = 2√2 b) AC = AE + EC AE/DE = tan60? = √3 AE = (√3)(DE) = 2√3 EC/DE = tan45? = 1 EC = DE = 2 AC = AE + EC = 2√3 + 2 c) AB/AC = cos45? = 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 = 90?, angle ACB = angle BAC = 45?. 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 = 75? DG = AB = (√2)(√3 + 1) 2/AD = cos60? = 1/2 AD = 4 sin75? = DG/AD = [(√2)(√3 + 1)]/4 = (1+√3)/(2√2)