Home » Ask & Discuss » Mathematics. » Trignometry « Back to Discussion

Trignometry

Blazing goIITian

Joined:	11 Jun 2009
Post:	386

18 Jun 2009 23:52:54 IST

0 People liked this

324

find the values of p for which the eq sinx+pcosx=2p has a solution

None

Share this article on:

Comments (3)

taran

Blazing goIITian

Joined: 17 Nov 2008

Posts: 1331

23 Jun 2009 12:54:52 IST

Like 0 people liked this

clearly 2p has to lie between max and min values of the equation i.e

Blazing goIITian

Joined: 11 Jun 2009

Posts: 386

23 Jun 2009 15:42:32 IST

Like 0 people liked this

hey dude you get values of p i.e they lie between (-3)^-0.5 and (3)^0.5

but how will we reach they answer

®µD®A

Blazing goIITian

Joined: 12 Apr 2008

Posts: 2717

24 Jun 2009 09:43:29 IST

Like 2 people liked this

Let sin x + pcos x = rsin(x+alpha)=rsin xcosalpha + rcos x sin alpha

So rcosalpha = 1 and rsinalpha = p

Suqring both and adding we get,

$r^2(cos^2alpha + sin^2alpha) = 1+p^2Rightarrow r = sqrt{1+p^2}$

So,

$sqrt{1+p^2}sin(x+alpha) = 2pRightarrow sin(x+alpha) = rac{2p}{sqrt{1+p^2}}$

This has a solution if $-1 le rac{2p}{sqrt{1+p^2}} le 1$

$rac{2p}{sqrt{1+p^2}} leq 1 Rightarrow 4p^2 leq 1+p^2 Rightarrow p leq rac{1}{sqrt 3}$

If p is negative,

$-p \leq\frac1{\sqrt 3} \Rightarrow p \geq -\frac{1}{\sqrt 3}$

So we have,

-rac{1}{sqrt 3} leq p leq rac{1}{sqrt 3}

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team

Free Sign Up!

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor

@ INR 1,995/-

For Quick Info

Find Posts by Topics

Physics.

Topics

9401

20093

3672

2186

7617

2827

2638

Mathematics.

Chemistry.

Biology

Institutes

Parents

Board

Fun Zone