solving Trinnometry

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1. Solve : 2 tan x = 4x - 2 sin 2x cos 2x tan² x

2. Show that the solutions of the equation tan ax = tan bx, a² + b² <> 0 are in A.P. Find the common difference of the series.

3. If α , β be the roots of the equation a cos 2θ + b sin 2θ =c then show that
tan α + tan β = 2b / c + a and tan α tan β = c - a / c + a

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@nishat,

Welcome to a2k.

We have an unofficial ban on doing people's homework here, but these problems look so difficult that I would like to know the answers as well. I have no idea how to do these and would love to hear others' thoughts, but here are my initial ones...

nishat wrote:

1. Solve : 2 tan x = 4x - 2 sin 2x cos 2x tan² x

Graphing the function f(x) = 4x - 2 sin 2x cos 2x tan² x - 2 tan x shows that f(x) has infinitely many zeroes, but only one zero on the interval [-π/2, π/2]. Do you need all the answers, or just one?

Quote:

2. Show that the solutions of the equation tan ax = tan bx, a² + b² <> 0 are in A.P. Find the common difference of the series.

Sorry to sound ignorant here, but (i) what is this question asking? (ii) what is "<>"? and (iii) what is meant by "A.P."?

Quote:

3. If α , β be the roots of the equation a cos 2θ + b sin 2θ =c then show that
tan α + tan β = 2b / c + a and tan α tan β = c - a / c + a

Well, you can parameterise a the unit circle as: (x,y) = (cos 2θ, sin 2θ); then find its points of intersection with the line ax + by = c. Draw a line through the origin and each point of intersection. The slopes of those lines will be (tan α) and (tan β).

Still, I'm not sure how you'd go about doing this.

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@Oylok,

Oylok wrote:

nishat wrote:

3. If α , β be the roots of the equation a cos 2θ + b sin 2θ =c then show that
tan α + tan β = 2b / c + a and tan α tan β = c - a / c + a

Well, you can parameterise a the unit circle as: (x,y) = (cos 2θ, sin 2θ); then find its points of intersection with the line ax + by = c. Draw a line through the origin and each point of intersection. The slopes of those lines will be (tan α) and (tan β).

Oops...

I think that should read:
"The slopes of those lines will be (tan 2α) and (tan 2β)."

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@Oylok,

Oylok wrote:

Sorry to sound ignorant here, but ... (ii) what is "<>"?

It means "greater or smaller" (the "different" sign isn't easy to print online). You can enter equations and get solutions on this link: http://www.wolframalpha.com/
by entering "trigonometry" (no quotes) in the search line - of course, if you write the word "trinnometry" (sic) you may not get the exact same result Smile

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Dear All,

All the questions are from West Bengal Council of Higher Secondary Examination. The symbol means not equal to (≠ ) and A.P. means Arithmetic Progression.

Thank for early reply.

Nishat

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@Oylok,

There isn't such a ban, either official or unofficial. Many people take that approach, for whatever reason. My own inclination is to answer when I know the answer, and remain silent when I don't.

Honestly, I'm always proud of myself when I even understand the question.

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@Oylok,

I'll be the one to argue with you, Oylok, much as I like you.

The person who owns this site is interested in information. He has remonstrated off and on here about people saying they won't do homework, not to speak for him. This is an information site, primarily.

As a culture, we tend to try to help people work out an answer.

Also as a culture, we tend to slam people asking about homework. As you did.

Pick the first one if you can. The second is unwarranted.

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@ossobuco,

ossobuco wrote:

As a culture, we tend to try to help people work out an answer.

Also as a culture, we tend to slam people asking about homework. As you did.

Pick the first one if you can. The second is unwarranted.

For my own part, I tend towards giving hints, and though mine usually lack the wit I've seen in the hints from some other Math wizards at this site, they hopefully do prove helpful. But I try to steer clear of spoon-feeding people complete answers.

When I was first in college, I tried to conquer on my own every problem I encountered, even while others collaborated and gained ground on me. I got into some nasty jams and would spend hours just staring at the textbook, because I was thinking about things the wrong way. I want to help others avoid those jams, and I don't see anything particularly wrong with playing the role of either a teacher or even a "schoolmate" here at a2k. (Which role I take will depend on how hard the question is. Smile

) The fact is, collaboration is part of Math, and usually when someone is stuck on problem, it's the case that some minor detail is tripping them up.

Cheers,
Oylok

PS--I have no clue who keeps voting your more critical replies to me down, but it's making me look bad. I usually fix your score when I see it inexplicable at zero. Rolling Eyes

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@Oylok,

She knows that. There is someone that just follows her around and gives her the thumb. It happens to others, too. I get the same treatment, except I usually have to disagree with two certain people.

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@Oylok,

Good reply, I surely agree.

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@Oylok,

I have a tail wind recently. Might be about politics or religion or ... who knows.

I do the thumbs up thing too when people get thumbed for just saying what they think. Once in a while I let it go as that may take up pages and shows that there is some loon out there. I'm probably odd in that I'll thumb up good arguments from different sides, thus probably confusing the statistically minded. I work to keep my fingers from thumbing down people I disagree with.

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@roger,

Right. I'm hoping my thumber is the one downing at least one other person I'm aware of here, as that would be a compliment.

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@nishat,

Hello
An answer to the third question:
a cos 2θ + b sin 2θ =c

Let write the equation in the following way:
a cos 2θ + b sin 2θ =a(cos^2 θ - sin^2 θ )+a*2*sinθ *cosθ =
c= c(sin^ θ + cos ^θ )

I use the identities that I think you know:
sin^ θ + cos ^θ =1

sin(2θ )=2sinθ cosθ

cos 2θ = cos^2 θ - sin ^θ

Now we can divide the both sides of the equation by the term
sinθ *cosθ

The equation gets the form:
a/tgθ -a*tgθ +2b=c*tgθ + c/tg θ

Now we multiply the both sides by tg θ

The equation gets the form:

a-a*tg^2θ +2b*tgθ =c*tg^2 θ +c

c*tg^2 θ + a*tg^2θ - 2b*tgθ +(c-a)=0

(c+a)*tg^2θ -2b*tgθ +(c-a)=0

(We assume that c+a !=0, otherwisw the equalities are not valid)

We substitute t=tg θ

A=c+a B=-2b C=c-a

We get

At^2+ B*t +C=0

This is a quadratic equation
t1+t2=-B/A
t1*t2=C/A

Let return back to the original substitutions

t1+t2= -(-2b)/(c+a)=2b/(c+a)

t1*t2=(c-a)/(c+a)

Now since α and β are the roots, therefore

t1=tan α; t2=tan β
and

tg α + tg β = 2b / c + a
and tg α tg β = c - a / c + a

I'll be happy to answer to questions:
[email protected]

Amos

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@AM1,

Nice answer.

Looks like the double-angle formulas were the way to go after all.

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@nishat,

nishat wrote:

2. Show that the solutions of the equation tan ax = tan bx, a² + b² <> 0 are in A.P. Find the common difference of the series.

After your explanation, this problem now appears to be the easiest of the three. The trick is that tan(-) is periodic, with period π. So that means ax and bx differ by n*π, where n is some integer. Don't know how much that helps...

π = PI (for those of you with sub-par web-browsers) Rolling Eyes

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solving Trinnometry

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