# Maths Question

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#1
solve :

2sin2x=tanx

for 0-360 deg
0
16 years ago
#2
(Original post by ShOcKzZ)
solve :

2sin2x=tanx

for 0-360 deg
Use the double angle formula first to simplify left hand side....

can't quite recall them...but its something like...

sin2x = 2sinxcosx????

G
0
16 years ago
#3
(Original post by gzftan)
Use the double angle formula first to simplify left hand side....

can't quite recall them...but its something like...

sin2x = 2sinxcosx????

G
Yeah, that's right.
0
#4
(Original post by Nylex)
Yeah, that's right.
yes, so:

2 [ 2sinxcosx ] = tanx

4sinxcosx = sinx/cosx

4sinxcos^2x = sinx

4sinx(1-sin^2x) = sinx

now??..
0
16 years ago
#5
(Original post by ShOcKzZ)
solve :

2sin2x=tanx

for 0-360 deg
Is this P3 Maths Shockzz, yes?
0
16 years ago
#6
(Original post by ShOcKzZ)
yes, so:

2 [ 2sinxcosx ] = tanx

4sinxcosx = sinx/cosx

4sinxcos^2x = sinx

4sinx(1-sin^2x) = sinx

now??..
4sinx(1-sin^2x) - sinx = 0
4sinx(1-sin^2x - 1/4) = 0.

So sin(x) = 0

or sin^2x = 3/4 so sin(x) = +/- root3/2.
0
#7
(Original post by bono)
Is this P3 Maths Shockzz, yes?
p4 aqa B
i think thats p3 work for most other boards.
0
16 years ago
#8
2sin(2x) = tan(x)

So, using the double angle formula:

sin(2x) = sin(x+x) = sin(x)cos(x)+sin(x)cos(x) = 2sin(x)cos(x)

=> 2(2sin(x)cos(x)) = tan(x)

We know that tan(x) = sin(x)/cos(x):

4sin(x)cos(x) = sin(x)/cos(x)

Cancel the sin(x):

4cos(x) = 1/cos(x)
4cos^2(x) = 1
cos^2(x) = 1/4
cos(x) = sqrt(1/4) = 1/2

So,

x = cos^-1(1/2) = 60 deg.
OR x = 360 - 60 = 300 deg.

I checked this numerically too.
0
#9
(Original post by theone)
4sinx(1-sin^2x) - sinx = 0
4sinx(1-sin^2x - 1/4) = 0.

So sin(x) = 0

or sin^2x = 3/4 so sin(x) = +/- root3/2.
amazing, i have repped you, thanks.
0
16 years ago
#10
(Original post by capslock)
2sin(2x) = tan(x)

So, using the double angle formula:

sin(2x) = sin(x+x) = sin(x)cos(x)+sin(x)cos(x) = 2sin(x)cos(x)

=> 2(2sin(x)cos(x)) = tan(x)

We know that tan(x) = sin(x)/cos(x):

4sin(x)cos(x) = sin(x)/cos(x)

Cancel the sin(x):

4cos(x) = 1/cos(x)
4cos^2(x) = 1
cos^2(x) = 1/4
cos(x) = sqrt(1/4) = 1/2

So,

x = cos^-1(1/2) = 60 deg.
OR x = 360 - 60 = 300 deg.

I checked this numerically too.
a) You can't cancel the sin(x) (what if they are 0?)

b) You've forgotten to take the negative root of 1/4 into account as well.
0
16 years ago
#11
(Original post by ShOcKzZ)
amazing, i have repped you, thanks.
Yes Shockzz I do the same syllabus!!

P1, P2, M1, P4, P5, P2.
0
16 years ago
#12
(Original post by ShOcKzZ)
amazing, i have repped you, thanks.
No prob, thanks 0
#13
(Original post by bono)
Yes Shockzz I do the same syllabus!!

P1, P2, M1, P4, P5, P2.
u do p2 twice?? hehe
0
16 years ago
#14
(Original post by ShOcKzZ)
p4 aqa B
i think thats p3 work for most other boards.
Yes, AQA's modules are a bit different.
0
#15
(Original post by bono)
Yes, AQA's modules are a bit different.
yes, i do

p1,p2,s1,p4,p5,m3
0
16 years ago
#16
(Original post by theone)
4sinx(1-sin^2x) - sinx = 0
4sinx(1-sin^2x - 1/4) = 0.
- sinx = -1/4 0
16 years ago
#17
(Original post by ShOcKzZ)
yes, i do

p1,p2,s1,p4,p5,m3
Why you gonna do M3 when you can do M2?

And don't you need M1 to do a later Mechanics module?? 0
16 years ago
#18
(Original post by bono)
- sinx = -1/4 Err, what? (1 - sin^2(x) - 1/4) = 0 <=> 3/4 - sin^2(x) = 0 <=> 3/4 = sin^2(x).
0
#19
(Original post by bono)
Why you gonna do M3 when you can do M2?

And don't you need M1 to do a later Mechanics module?? I have no idea.
0
16 years ago
#20
How the hell could AQA possibly justify leaving double angle formula off P3?
0
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