# Maths Question

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#2

(Original post by

solve :

2sin2x=tanx

for 0-360 deg

**ShOcKzZ**)solve :

2sin2x=tanx

for 0-360 deg

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

sin2x = 2sinxcosx????

G

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#3

(Original post by

Use the double angle formula first to simplify left hand side....

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

sin2x = 2sinxcosx????

G

**gzftan**)Use the double angle formula first to simplify left hand side....

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

sin2x = 2sinxcosx????

G

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(Original post by

Yeah, that's right.

**Nylex**)Yeah, that's right.

2 [ 2sinxcosx ] = tanx

4sinxcosx = sinx/cosx

4sinxcos^2x = sinx

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

now??..

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#5

(Original post by

solve :

2sin2x=tanx

for 0-360 deg

**ShOcKzZ**)solve :

2sin2x=tanx

for 0-360 deg

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#6

(Original post by

yes, so:

2 [ 2sinxcosx ] = tanx

4sinxcosx = sinx/cosx

4sinxcos^2x = sinx

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

now??..

**ShOcKzZ**)yes, so:

2 [ 2sinxcosx ] = tanx

4sinxcosx = sinx/cosx

4sinxcos^2x = sinx

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

now??..

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

So sin(x) = 0

or sin^2x = 3/4 so sin(x) = +/- root3/2.

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(Original post by

Is this P3 Maths Shockzz, yes?

**bono**)Is this P3 Maths Shockzz, yes?

i think thats p3 work for most other boards.

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#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.

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.

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(Original post by

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.

**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.

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#10

(Original post by

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.

**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.

b) You've forgotten to take the negative root of 1/4 into account as well.

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#11

(Original post by

amazing, i have repped you, thanks.

**ShOcKzZ**)amazing, i have repped you, thanks.

P1, P2, M1, P4, P5, P2.

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(Original post by

Yes Shockzz I do the same syllabus!!

P1, P2, M1, P4, P5, P2.

**bono**)Yes Shockzz I do the same syllabus!!

P1, P2, M1, P4, P5, P2.

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#14

(Original post by

p4 aqa B

i think thats p3 work for most other boards.

**ShOcKzZ**)p4 aqa B

i think thats p3 work for most other boards.

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(Original post by

Yes, AQA's modules are a bit different.

**bono**)Yes, AQA's modules are a bit different.

p1,p2,s1,p4,p5,m3

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#16

(Original post by

4sinx(1-sin^2x) - sinx = 0

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

**theone**)4sinx(1-sin^2x) - sinx = 0

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

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#17

(Original post by

yes, i do

p1,p2,s1,p4,p5,m3

**ShOcKzZ**)yes, i do

p1,p2,s1,p4,p5,m3

And don't you need M1 to do a later Mechanics module??

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#18

(Original post by

- sinx = -1/4

**bono**)- sinx = -1/4

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(Original post by

Why you gonna do M3 when you can do M2?

And don't you need M1 to do a later Mechanics module??

**bono**)Why you gonna do M3 when you can do M2?

And don't you need M1 to do a later Mechanics module??

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