# P3 OCR help plzWatch

This discussion is closed.
#1
could anyone have a look at this question its part 3 of the jan 2000 MEI paper:

f(x)=2(2cosx-sinx)sinx
when x=a f(x)=0
show that tan a=2
find the exact value of sin a
0
14 years ago
#2
taken from the lovely Meiklerigg's website:-
0
#3
ok I had done it this way:
2(2cos x-sinx)sinx=0 i divided by sinx and then 2
2cosx-sinx =0 i divided by cos x
2=tanx

then for the next bit
2=sinx/cosx
sinx =2cos x
sinx^2=4cosx^2
sinx^2=4(1-sinx^2)
sinx^2=4-4sinx^2
sinx^2(1+4)=4
sinx^2=4/5
sinx=square root of 4/5
sinx=2/square root of 5

where have I gone wrong?
0
14 years ago
#4
(Original post by *Katherine*)
ok I had done it this way:
2(2cos x-sinx)sinx=0 i divided by sinx and then 2
2cosx-sinx =0 i divided by cos x
2=tanx

then for the next bit
2=sinx/cosx
sinx =2cos x
sinx^2=4cosx^2
sinx^2=4(1-sinx^2)
sinx^2=4-4sinx^2
sinx^2(1+4)=4
sinx^2=4/5
sinx=square root of 4/5
sinx=2/square root of 5

where have I gone wrong?
looks right to me, just multiply top and bottom by rt5.
i.e. 2/rt5=2rt5/(rt5.rt5)
=2rt5/5
0
#5
yay so tomorrow isn't such a lost cause after all!! thanx for your help both of you.
0
14 years ago
#6
2rt5 / 5 = 2/rt5
0
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