Find the value (1 Viewer)

[Rkiuyto]

1. Find the value of tan²1+ tan²3+.. ...+tan² 89.

2.Find the ordered pairs x and y where x and y are real numbers such that x and y satisfy the two equations below.

(1/sqrt x)+(1/sqrt y)=(x+3y)(3x+y)
(1/sqrt x)-(1/sqrt y)=2 (y^2-x^2)

 

[Iruka]

For Q2, any pair of real numbers with x=y (except x=y=0) will satisfy the second of your simultaneous equations. So substitute x=y into the first equation, and you should find that x=y= 2^-6/5 is a solution.

 

[shaon0]

1. Find the value of tan²1+ tan²3+.. ...+tan² 89.

2.Find the ordered pairs x and y where x and y are real numbers such that x and y satisfy the two equations below.

(1/sqrt x)+(1/sqrt y)=(x+3y)(3x+y)
(1/sqrt x)-(1/sqrt y)=2 (y^2-x^2)

1.
=((cot(89))^2+(tan89)^2)+((cot(87))^2+(tan(87))^2)+...+((cot(43))^2+(tan(43))^2)+(tan45)^2

Does the approach above work?

 

[Timothy.Siu]

1.
=((cot(89))^2+(tan89)^2)+((cot(87))^2+(tan(87))^2)+...+((cot(43))^2+(tan(43))^2)+(tan45)^2

Does the approach above work?

and then what?

 

[bored of sc]

1. Find the value of tan²1+ tan²3+.. ...+tan²89

tan²x = sin²x/cos²x = sec²x/cosec²x = (1+tan²x)/(1+cot²x)

maybe change tan²x to cot²(90-x) for half of them (i.e. up to 43^o) and use cot²x = (1+cot²x)/(1+tan²x).

tan²1 = sin²1/cos²1 = cos²89/sin²89

sin²89 = 1-cos²89

WORKING starts here:
sin²1/sin²89 + sin²3/sin²87
+ ... + sin²43/sin²47 + 1 + cos²43/cos²47 + ... + cos²3/cos²87 + cos²1/cos²89

common denominator (without the 1 in it) is sin²47.cos²47*sin²49*cos²49*... ... sin²89*cos²89

numerator becomes 44*sin²1*cos²1*...*sin²43*cos²43

change denominator using complementary angles i.e. sin²47 = cos²43

thus everything cancels and we are left with 44 + 1 = 45

Listen to Trebla.

 

[Trebla]

tan²x = sin²x/cos²x = sec²x/cosec²x = (1+tan²x)/(1+cot²x)

maybe change tan²x to cot²(90-x) for half of them (i.e. up to 43^o) and use cot²x = (1+cot²x)/(1+tan²x).

tan²1 = sin²1/cos²1 = cos²89/sin²89

sin²89 = 1-cos²89

WORKING starts here:
sin²1/sin²89 + sin²3/sin²87
+ ... + sin²43/sin²47 + 1 + cos²43/cos²47 + ... + cos²3/cos²87 + cos²1/cos²89

common denominator (without the 1 in it) is sin²47.cos²47*sin²49*cos²49*... ... sin²89*cos²89

numerator becomes 44*sin²1*cos²1*...*sin²43*cos²43

change denominator using complementary angles i.e. sin²47 = cos²43

thus everything cancels and we are left with 44 + 1 = 45

Here's hoping that is right...

Your numerator is wrong. They will not have a common numerator.

 

[bored of sc]

Your numerator is wrong. They will not have a common numerator.

Damn, thanks. Haha, I don't know what I was thinking.

 

[Trebla]

Is the answer 4005?

 

[Rkiuyto]

Is the answer 4005?

yes

very good but how

 

[lyounamu]

Is the answer 4005?

yes

very good but how

Manual calculation FTW!

 

[shaon0]

and then what?

I cbf doing it. Might do it later if it's not solved.

 

[lyounamu]

I cbf doing it. Might do it later if it's not solved.

well done, that's as good as manual calulation.

 

[shaon0]

well done, that's as good as manual calulation.

I'm a bit busy atm with other stuff. No point of doing extra work on top of stuff i need to do for school.

 

[Trebla]

Manual calculation FTW!

Yep that's exactly what I did because I couldn't do anything else!!!

That's not to say I haven't tried getting it without just manual calculation. I even tried something crazy like bound the series with Riemann sums of y = tan²x (approximations to integrals by upper and lower rectangles) to form a two sided inequality to get at least a good approximation but that totally failed...lol

Mind you, it's quite an elegant result...

 

[lolokay]

haha
Rkiuyto, where is this question (the tan² one) from?

solution is here.. Art of Problem Solving Forum

 

[shaon0]

1. Find the value of tan²1+ tan²3+.. ...+tan² 89.

2.Find the ordered pairs x and y where x and y are real numbers such that x and y satisfy the two equations below.

(1/sqrt x)+(1/sqrt y)=(x+3y)(3x+y)
(1/sqrt x)-(1/sqrt y)=2 (y^2-x^2)

Rkiuyto, You do realise this is almost Pre-Olympiad Mathematics.
You shouldn't really expect many people on BoS to get it.

 

[Timothy.Siu]

lol, thats crazy...i sorta got lost

 

[Trebla]

Complex numbers and Polynomials...in a 2 unit forum...lol

Even in an Ext2 paper, there would be a lead in with the (x + i)/√(x² + 1) bit

 

[Iruka]

That proof was interesting, though. I've seen stuff like that using cyclotomic polynomials with cos and sin, but not cot and tan.

One of the chapters in Hardy's A Course of Pure Mathematics has a whole heap of questions like that.