f.5 Trigonometry

LHS=[2sin²(90-x)-1]/[sin(90-x)+sinx]
=[2cos²x-(sin²x+cos²x)]/(cosx+sinx)
=(cos²x-sin²x)/(cosx+sinx)
=[(cosx+sinx)(cosx-sinx)]/(cosx+sinx)
=cosx-sinx
=RHS

2011-08-22 21:16:02 補充：
can u type it again more clearly?

2011-08-22 21:17:33 補充：
?? '20tan²' ??

2011-08-23 21:37:19 補充：
??20tan^2???

do u mean 20tan²x??

2011-08-24 13:12:28 補充：
(a).
[2sin²(90-x)-1]/[sin(90-x)+sinx]
=(2cos²x-1)/(cosx+sinx)
=[cos²x+(cos²x-1)]/(cosx+sinx)
=(cos²x-sin²x)/(cosx+sinx)
=[(cosx+sinx)(cosx-sinx)]/(cosx+sinx)
=cosx-sinx

So,
cosx-sinx=20sinxtanx
cosx=20sinxtanx+sinx
1=20tan²x+tanx
20tan²x+tanx-1=0

2011-08-24 13:15:47 補充：
(b).
[2sin²(90-x)-1]/[sin(90-x)+sinx]=20sinxtanx
From(a),
20tan²x+tanx-1=0
By quadratic formula,
tanx=0.2 or -0.25
So,
x=11.3,191 or 166,346(3 sig.fig.)

Provethat ：
2Sin^2 (90°－x)－1
----------------------
Sin(90°－x)+Sinx
=Cosx－Sinx
Sol
2Sin^2(90°－x)－1=2cos^2 x－1=Cos^2x－Sin^2 x
Sin(90°－x)+Sinx=Cosx+Sinx
So
2Sin^2 (90°－x)－1 　Cos^2 x－Sin^2 x
------------------------=-----------------------------=Cosx－Sinx
Sin(90°－x)+Sinx Cosx+Sinx