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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Inside Out on April 16, 2011, 07:36:31 pm

Title: TRIGALICOUS
Post by: Inside Out on April 16, 2011, 07:36:31 pm
 :D  For (pie/2)<x<pie with cosa=-sinb where o<b<(pie/2), find a in terms of b





Title: Re: TRIGALICOUS
Post by: BubbleWrapMan on April 16, 2011, 07:52:55 pm
*noms trig*

These questions love diagrams. Try it, then report back.
Title: Re: TRIGALICOUS
Post by: Water on April 16, 2011, 07:54:25 pm
(pie/2)<x<pie with cosa=-sinb where o<b<(pie/2)


cos a = -sin b



sin (pi/2 + a ) = sin (b + pi)

pi / 2 + a = b + pi

           a = b + pi/2

           a =  (b + pi/2)



Not sure if this is right, can I get it confirmed :)?

*Corrected*


        
Title: Re: TRIGALICOUS
Post by: Inside Out on April 16, 2011, 07:56:08 pm
answer is : b + pi/2
Title: Re: TRIGALICOUS
Post by: Water on April 16, 2011, 08:01:34 pm
cos a =   - sin b

        =    cos ( pi/2 + b)


           a = pi/2 + b


I'll check with other, dunno why it doesn't work ):
   


Title: Re: TRIGALICOUS
Post by: Inside Out on April 16, 2011, 08:04:07 pm
ok thanks :)
Title: Re: TRIGALICOUS
Post by: Inside Out on April 16, 2011, 08:14:15 pm
how do you know whether cosa=sin (pi/2 - a ) or sin (pi/2 + a )
Title: Re: TRIGALICOUS
Post by: Water on April 16, 2011, 08:22:52 pm
Complementary angles, two ways to learn it, either graph it out and understand theory, or rote learn it. :), Yeah sorry, fixed my error before xD
Title: Re: TRIGALICOUS
Post by: Inside Out on April 16, 2011, 08:25:20 pm
yes but can't cosa= either sin (pi/2 + a )  or sin (pi/2 -a ).. how did u know which one to chose?
Title: Re: TRIGALICOUS
Post by: enpassant on April 16, 2011, 08:33:28 pm
For (pie/2)<x<pie
Title: Re: TRIGALICOUS
Post by: Inside Out on April 16, 2011, 08:57:27 pm
yep i get that now :)...
but how come -sinb cant equal sin(2pi-b) instead of sin(pi+b)
Title: Re: TRIGALICOUS
Post by: enpassant on April 16, 2011, 11:19:51 pm
Quote from: happyface on April 16, 2011, 08:57:27 pm
yep i get that now :)...
but how come -sinb cant equal sin(2pi-b) instead of sin(pi+b)
-sinb cant equal sin(2pi-b) because this will give a=3pi/2 - b which is not in the second quad as required ((pie/2)<x<pie)
Title: Re: TRIGALICOUS
Post by: Water on April 16, 2011, 11:20:56 pm
 o<b<(pie/2)



 sin (b + pi)


Wouldn't this therefore be wrong, as this will be not be in the 1st quad? My method of figuring this out could be completely wrong , I'm doubting  ):!
Title: Re: TRIGALICOUS
Post by: enpassant on April 16, 2011, 11:38:20 pm
Quote from: Water on April 16, 2011, 11:20:56 pm
o<b<(pie/2)



 sin (b + pi)


Wouldn't this therefore be wrong, as this will be not be in the 1st quad? My method of figuring this out could be completely wrong , I'm doubting  ):!

symmetry property: -sinb=sin(b+pi)=sin(2pi-b)
b in the first quad, b+pi in the third quad and 2pi-b in the fourth quad
-sinb cant equal sin(2pi-b) because this will give a=3pi/2 - b which is not in the second quad as required  by (pi/2)<a<pi
Title: Re: TRIGALICOUS
Post by: Water on April 16, 2011, 11:41:08 pm
Quote from: enpassant on April 16, 2011, 11:38:20 pm
Quote from: Water on April 16, 2011, 11:20:56 pm
o<b<(pie/2)



 sin (b + pi)


Wouldn't this therefore be wrong, as this will be not be in the 1st quad? My method of figuring this out could be completely wrong , I'm doubting  ):!

symmetry property: -sinb=sin(b+pi)=sin(2pi-b)
b in the first quad, b+pi in the third quad and 2pi-b in the fourth quad
-sinb cant equal sin(2pi-b) because this will give a=3pi/2 - b which is not in the second quad as required  by (pi/2)<a<pi


excellent explanation :) You are awesome xD
Title: Re: TRIGALICOUS
Post by: Inside Out on April 17, 2011, 01:02:24 pm
yay i get this now.. thanks :)