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**debjit625****Member**- Registered: 2012-07-23
- Posts: 91

Well let me tell this first,I am not asking for any solution as I have them ,my problem is something different...

I can do trigonometric prove but only to a limit for example these kind of questions I have no problem

or

or

or more general types I have no problems ,but there are some I am having problems and I understand that, I am not understanding the process or steps involved in solving them for example

Question 1

Question 2

I am not able to solve these kind problems ,what I need to know is what steps you guys take to solve these kind of problems,actually I feel them very complex.

In short I do not need the solution for question 1 and 2 I need to know how you guys solve these.

Another thing ,my book have very few of these questions so I can't even practice more if you guys have any resource it would be great.

Thanks

*Last edited by debjit625 (2013-05-05 21:12:53)*

Debjit Roy

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The essence of mathematics lies in its freedom - Georg Cantor

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,273

hi Debjit Roy

Well they look very complex to me, too. And I haven't a 'method' for either. So I'll just tell you where my thoughts have gone. Maybe it will help; maybe someone else will come up with a 'slick' solution.

Q1. z is the only 'unknown here so it should be possible to eliminate it, at the cost of one equation, leaving a single equation that leads to the required form.

I took the second equation and wrote sin(z) in terms of cos(z). Then I substituted into the first equation, hoping all the cos terms would cancel. Huh! No such luck!

But you do get a single cos(z) term so maybe you could substitute that back along with the expression for sin(z). But I fear that might lead to everything cancelling and 0 = 0. Re-assuring, but not much help!

But we also know that sin squared plus cos squared makes 1, so you could get a second expression for cos(z) and then eliminate it. This 'must' leave the required equation, since it is a valid equation with all the right algebraic terms. But It looks like a very messy lot of algebraic simplification will still be needed.

I'll keep thinking about this.

Q2. This time we have two unknowns to eliminate ( p and q) and three equations. So, again it should be possible. If the incentive was strong enough (eg. I was offered a lot of money) I could use the quadratic formula on each of the first and second equations to get expressions for q and for p and then substitute these back into the third equation. Once again it looks horribly messy, especially as the quad. formula has two solutions for each of q and p. Yuck!

I wonder what happens if you use the third to make cos the subject and then sub back into the other equations. You don't really want to eliminate the cos, but maybe it makes the equations simpler.

Or maybe take the third, square it and then try to eliminate p and q that way. I think I'll give that a try now. Back if it works.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**debjit625****Member**- Registered: 2012-07-23
- Posts: 91

Thanks Bob for the reply

Debjit Roy

___________________________________________________

The essence of mathematics lies in its freedom - Georg Cantor

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