Given that 2sin2x + cos2x = a, show that (1 + a)tan^2x-4tanx+a=1

[baked]

Hi Everyone,
This is the question I don't understand

a) Given that 2sin2x + cos2x = a, show that (1 + a)tan²x-4tanx+a=1
b) Hence or otherwise, show that if tan x1 and tan x2 are the roots of the quadratic in (a) then tan (x1 + x2) = 2

This is what I have attempted to do for part a, I don't really know how to proceed or if I missed a previous step.

(4sinxcosx + 2cos²x)(tan²x) - 4tanx + (4sinxcosx + 2cos²x - 1)
= 4sin³x/cosx + 2sin²x - 4tanx + (4sinxcosx + 2cos²x - 1)
= 4sin(1-cos²x)/cosx + 2sin²x - 4tanx + 4sinxcosx + 2cos²x - 1
= 4sin(1-cos²x)/cosx - 4tanx + 4sinxcosx + 2sin²x + 2cos²x - 1
= 4sin(1-cos²x)/cosx - 4tanx + 4sinxcosx + 2(1) - 1
= 4sin(1-cos²x)/cosx - 4tanx + 4sinxcosx + 1

thx

 

[Cubist]

Hi, welcome to the forum!

You made a shortcut in your notation that might be confusing you (it certainly confused me)

= 4sin³x/cosx + 2sin²x - 4tanx + (4sinxcosx + 2cos²x - 1)
= 4sin(1-cos²x)/cosx + 2sin²x - 4tanx + 4sinxcosx + 2cos²x - 1

4sin³x ≠ 4sin(1-cos²x)

4sin³(x) = 4sin(x)*(1 - cos²(x))

Correct this on your last line and then you're close to completing part (a). Distribute (expand) the bracket and things will cancel

 

[baked]

Hi, welcome to the forum!

You made a shortcut in your notation that might be confusing you (it certainly confused me)


4sin³x ≠ 4sin(1-cos²x)

4sin³(x) = 4sin(x)*(1 - cos²(x))

Correct this on your last line and then you're close to completing part (a). Distribute (expand) the bracket and things will cancel

Hi there!
Thank you very much, I see how that made it very confusing. Distributed and got there in the end
4sinx(1-cos²x)/cosx - 4sinx/cosx+ 4sinxcosx + 1
= (4sinx(1-cos²x) - 4sinx + 4sinxcos²x + cosx ) / cosx
= 4sinx - 4sincos²x - 4sinx + 4sinxcos²x + cosx) / cosx
= cosx/cosx
= 1

For part (b), here is where I have started and was wondering if it is in the right direction?
I didn't directly use part a's result, but do I need to?

tan x1 + tan x2 = 4 / a + 1
(tanx1)*(tanx2) = (a-1)/(a+1)

tan (x1 + x2) = (tan x1 + tan x2)/(1 - (tanx1)*(tanx2))
tan (x1 + x2) = (4 / a + 1) / (1 - (a-1)/(a+1))
= (4 / a+1) / (2 / a+1)
= 4 / 2
= 2

 

[Cubist]

Hi there!
Thank you very much, I see how that made it very confusing. Distributed and got there in the end
4sinx(1-cos²x)/cosx - 4sinx/cosx+ 4sinxcosx + 1
= (4sinx(1-cos²x) - 4sinx + 4sinxcos²x + cosx ) / cosx
= 4sinx - 4sincos²x - 4sinx + 4sinxcos²x + cosx) / cosx
= cosx/cosx
= 1

Well done! FYI: This is how I did it...

4sin(x)(1-cos²x)/cos(x) - 4tan(x) + 4sin(x)cos(x) + 1
=4sin(x)/cos(x) - 4sin(x)cos(x) - 4tan(x) + 4sin(x)cos(x) + 1

=4sin(x)/cos(x) - 4sin(x)cos(x) - 4tan(x) + 4sin(x)cos(x) + 1
=4sin(x)/cos(x) - 4tan(x) + 1
=1

For part (b), here is where I have started and was wondering if it is in the right direction?
I didn't directly use part a's result, but do I need to?

tan x1 + tan x2 = 4 / a + 1
(tanx1)*(tanx2) = (a-1)/(a+1)

tan (x1 + x2) = (tan x1 + tan x2)/(1 - (tanx1)*(tanx2))
tan (x1 + x2) = (4 / a + 1) / (1 - (a-1)/(a+1))
= (4 / a+1) / (2 / a+1)
= 4 / 2
= 2

You don't have to use the part "a" result because the question was, "Hence or otherwise" and you chose the "otherwise" option - and it works well Please use brackets to make your intent more clear, since

4/a+1 ≠ 4/(a+1)

I've added brackets to your work for anyone else viewing this thread...

tan x1 + tan x2 = 4 / (a + 1)
(tanx1)*(tanx2) = (a-1)/(a+1)

tan (x1 + x2) = (tan x1 + tan x2)/(1 - (tanx1)*(tanx2))
tan(x1 + x2) = (4 / (a + 1)) / (1 - (a-1)/(a+1))
= (4 / (a+1)) / (2 / (a+1))
= 4 / 2
= 2

FYI: I can't immediately see how to easily use the part "a" result ?‍

 

[baked]

Well done! FYI: This is how I did it...

4sin(x)(1-cos²x)/cos(x) - 4tan(x) + 4sin(x)cos(x) + 1
=4sin(x)/cos(x) - 4sin(x)cos(x) - 4tan(x) + 4sin(x)cos(x) + 1

=4sin(x)/cos(x) - 4sin(x)cos(x) - 4tan(x) + 4sin(x)cos(x) + 1
=4sin(x)/cos(x) - 4tan(x) + 1
=1

Thank you very much!! It's good to know there's another way to do it, I think your method was definitely more simplified than mine.

You don't have to use the part "a" result because the question was, "Hence or otherwise" and you chose the "otherwise" option - and it works well Please use brackets to make your intent more clear, since

4/a+1 ≠ 4/(a+1)

I've added brackets to your work for anyone else viewing this thread...

tan x1 + tan x2 = 4 / (a + 1)
(tanx1)*(tanx2) = (a-1)/(a+1)

tan (x1 + x2) = (tan x1 + tan x2)/(1 - (tanx1)*(tanx2))
tan(x1 + x2) = (4 / (a + 1)) / (1 - (a-1)/(a+1))
= (4 / (a+1)) / (2 / (a+1))
= 4 / 2
= 2

FYI: I can't immediately see how to easily use the part "a" result ?‍

Apologies, I'm not very used to typing out equations like this but I will make sure to keep brackets organised in the future. I spent a long time trying to figure out how I might do it from part a, but couldn't and had to resort to "otherwise" haha.

Thanks again for all of your help!

 

[Cubist]

Here's a way to show (b) using (a)

Using (a) we have 2*sin(2*x1) + cos(2*x1) = a = 2*sin(2*x2) + cos(2*x2)
2*sin(2*x1) - 2*sin(2*x2) + cos(2*x1) - cos(2*x2) = 0

Apply difference of cosines and difference of sines to obtain...

4*sin(x1-x2)*cos(x1+x2) - 2*sin(x1-x2)*sin(x1+x2) ) = 0
2 * sin(x1-x2) * ( 2*cos(x1+x2) - sin(x1+x2) ) = 0
sin(x1-x2) * ( 2 - tan(x1+x2) ) = 0

Therefore sin(x1-x2) = 0 or tan(x1+x2) = 2