Trigonometry questions

• bonkysleuth

  22 Jun 09, 12:30

   

  Can't see the pattern in this question. I guess there's supposed to be another theta sign lying around somewhere. Would help greatly if someone could help me out with this question.

  As for 15, I again, can't see the theta sign.For this question, I would just want someone to point out to me where the "other theta" sign could be. I want to attempt the rest of the question on my own. Thanks.

   

  If cosec x - sin x = a^3 and sec x - cos x = b^3, then simplify the expression a^2b^2(a^2 + b^2) + 9. (Hint: the answer should be a whole number)

   

  Lastly, if sin x + sin y = p, cos x + cos y = q and theta is the mean of x and y then find sin 2(theta).

  Which means you are required to find sin (x+y). I can get the answer, it is 2pq/ (p^2 + q^2) but only because this is a MCQ question.

  Please help, thanks!

• eagle

  22 Jun 09, 14:04

  I can't see photobucket images at the moment, so I shall leave it to other helpful forumers like Mr Wee and ForbiddenSinner to help you for that

   

  If cosec x - sin x = a^3 and sec x - cos x = b^3, then simplify the expression a^2b^2(a^2 + b^2) + 9. (Hint: the answer should be a whole number)

  we can change cosec x - sin x to sin²x/cos x and sec x - cos x to cos² x /sin x

  also, expand out a²b²(a² + b²)

  to get a^4 from a³, just take (a³)^4/3, which is (sin²x/cos x)^4/3

  Do it for the other term

  You will get a single trigo which you should be able to simplify. But I didn't go and complete this out. Try it out and show me where you are stucked.

   

  Lastly, if sin x + sin y = p, cos x + cos y = q and theta is the mean of x and y then find sin 2(theta).

  Which means you are required to find sin (x+y). I can get the answer, it is 2pq/ (p^2 + q^2) but only because this is a MCQ question.

  Tip:

  pq = (sin x + sin y)(cos x + cos y) ===> somewhere along the way, you will get sin² x + cos² x which gives you 1

  Try it too :D

   

• MasterMoogle

  22 Jun 09, 14:14

  Originally posted by eagle:

  I can't see photobucket images at the moment, so I shall leave it to other helpful forumers like Mr Wee and ForbiddenSinner to help you for that

   

  we can change cosec x - sin x to sin²x/cos x and sec x - cos x to cos² x /sin x

  also, expand out a²b²(a² + b²)

  to get a^4 from a³, just take (a³)^4/3, which is (sin²x/cos x)^4/3

  Do it for the other term

  You will get a single trigo which you should be able to simplify. But I didn't go and complete this out. Try it out and show me where you are stucked.

   

  Tip:

  pq = (sin x + sin y)(cos x + cos y) ===> somewhere along the way, you will get sin² x + cos² x which gives you 1

  Try it too :D

   

  my lao shi still haven't got fully well yet, so dun think he will come online

  

  

  lai lai, come and help out more

  edit: jialat, still the same need someone put on tinypic or something like dat

• eagle

  22 Jun 09, 14:23

  Pictures from xs.to not blocked though

  can upload there

• MasterMoogle

  22 Jun 09, 14:34

  Originally posted by eagle:

  Pictures from xs.to not blocked though

  can upload there

  http://xs140.xs.to/xs140/09261/amathsqn916.png

  http://xs140.xs.to/xs140/09261/amathsquestion14781.png

  upload hao liao. lao yin kuai dian lai zhuo gong ker

• wee_ws

  22 Jun 09, 15:16

  Hi,

    For amathsqn916.png:

    1. angle ABC = 90 degrees, since AC is a diameter.

    2. So angle ADC = 90 degrees.

    3. Drop a perpendicular from B to DC. Call it X.

    4. Then angle ABX = theta.

    Now you should be able to find AD.

    Thanks!

  Cheers,
  Wen Shih

• bonkysleuth

  23 Jun 09, 09:00

  Anyone has got any idea how to solve this question?

  http://xs140.xs.to/xs140/09261/amathsquestion14781.png

  Thanks. Haven't solved all the questions yet but will post up what I have done after I completed the questions

• bonkysleuth

  23 Jun 09, 10:26

  Hi eagle,

  For this earlier question I asked :

  If cosec x - sin x = a^3 and sec x - cos x = b^3, then simplify the expression a^2b^2(a^2 + b^2) + 9. (Hint: the answer should be a whole number)

  I found that the answer is 10. Managed to simplify the equation of (cos^4 x /cos x) + (sin^4 x /sin x ) = 1

  I found that the equation is equal to 1 using the calculator but I can't show how you derive that. Can you please help me out with this?

   

  And hi Mr Wee,

  I do not get why by dropping a perpendicular distance from B to C proves that angle ABX is theta.... hmm. Thanks for the help by the way! (and you may find this annoying, but I still can't see how to get distance of AD after that) >.<

   

  thanks all

• Forbiddensinner

  23 Jun 09, 11:52

  Originally posted by bonkysleuth:

  Hi eagle,

  For this earlier question I asked :

  If cosec x - sin x = a^3 and sec x - cos x = b^3, then simplify the expression a^2b^2(a^2 + b^2) + 9. (Hint: the answer should be a whole number)

  I found that the answer is 10. Managed to simplify the equation of (cos^4 x /cos x) + (sin^4 x /sin x ) = 1

  I found that the equation is equal to 1 using the calculator but I can't show how you derive that. Can you please help me out with this?

   

  And hi Mr Wee,

  I do not get why by dropping a perpendicular distance from B to C proves that angle ABX is theta.... hmm. Thanks for the help by the way! (and you may find this annoying, but I still can't see how to get distance of AD after that) >.<

   

  thanks all

  Are you sure the bolded equation is correct?

   

  angle BXC = 90 degree

  => angle XBC = ( 90 - theta ) degree

  angle ABC = 90 degree

  => angle ABX = angle ABC - angle XBC = 90 - ( 90 - theta ) degree = theta

   

• wee_ws

  23 Jun 09, 12:28

  Hi,

    Forbiddensinner already explained why angle ABX = theta.

    Now we can see that triangle BCX is a right angled triangle. Find BX.

    Drop another perpendicular from A to BX, call this point Y. Triangle ABY is another right angled triangle. Find BY.

    Now can you see any relationship between AD, BY and BX?

    Thanks!

  Cheers,
  Wen Shih

  P.S. Seriously, I feel that these questions are way too convoluted and they defeat the student's self-esteem rather than educate him/her the real concepts. You may wish to give feedback to the setters to provide hints, like dropping perpendiculars. Thanks!

• bonkysleuth

  23 Jun 09, 18:51

  Originally posted by Forbiddensinner:

  Are you sure the bolded equation is correct?

   

  angle BXC = 90 degree

  => angle XBC = ( 90 - theta ) degree

  angle ABC = 90 degree

  => angle ABX = angle ABC - angle XBC = 90 - ( 90 - theta ) degree = theta

   

  Thanks everyone. Managed to get the answer for this question.

  Hmm, regarding the equation I have given, I missed out on something.

  The equation should be

   (cos^4 x /cos x)^2/3 + (sin^4 x /sin x ) ^2/3= 1

  However, I do not know how to get 1 from that equation... and yea need help regarding the other question with the diagram. Thanks

• Forbiddensinner

  23 Jun 09, 19:13

  Originally posted by bonkysleuth:

  Thanks everyone. Managed to get the answer for this question.

  Hmm, regarding the equation I have given, I missed out on something.

  The equation should be

   (cos^4 x /cos x)^2/3 + (sin^4 x /sin x ) ^2/3= 1

  However, I do not know how to get 1 from that equation... and yea need help regarding the other question with the diagram. Thanks

  (cos^4 x /cos x)^2/3 + (sin^4 x /sin x ) ^2/3

  = (cos^3 x)^2/3 + (sin^3 x)^2/3

  = (cos^2 x) + (sin^2 x)

  = 1

• Audi

  27 Jun 09, 23:40

  Originally posted by bonkysleuth:

  Anyone has got any idea how to solve this question?

  http://xs140.xs.to/xs140/09261/amathsquestion14781.png

  Thanks. Haven't solved all the questions yet but will post up what I have done after I completed the questions

  
  Hi,

   

  Looks like no one else has yet answered this question of yours. I have done this question before but I can't manage to solve it now. I will go check up to see if I still have the worksheet. Meanwhile, maybe someone else might come to your help? Haha.

   

   

• Forbiddensinner

  28 Jun 09, 11:01

  Originally posted by Audi:

  
  Hi,

   

  Looks like no one else has yet answered this question of yours. I have done this question before but I can't manage to solve it now. I will go check up to see if I still have the worksheet. Meanwhile, maybe someone else might come to your help? Haha.

   

   

  The mods, me, and the other regulars here are stumped by this question.

  If we have even the slightest idea, we will have posted it up here already.

• Lee012lee

  28 Jun 09, 12:18

  Originally posted by Forbiddensinner:

  The mods, me, and the other regulars here are stumped by this question.

  If we have even the slightest idea, we will have posted it up here already.

  

  This question is about R-formula

  Steps :

  (1)   Let angle ABO be angle a.

         Let angle DBO be theta - a

  (2)  Let OB = b

        So, cos (theta - a) = z/b

  (3)  cos a = x/5b

        sin a = y/5b

  (4)  cos (theta - a) = cos theta cos a + sin theta sin a

  (5)  Substitute cos (theta - a) = z/b, cos a = x/5b, sin a = y/5b into

        cos (theta - a) = cos theta cos a + sin theta sin a

                 z/b         = cos theta (x/5b) + sin theta (y/5b)

                 z            = 1/5 [ x cos theta + y sin theta ]

  So, the answer is choice 3

   

• MasterMoogle

  28 Jun 09, 13:25

  Originally posted by Lee012lee:

  

  This question is about R-formula

  Steps :

  (1)   Let angle ABO be angle a.

         Let angle DBO be theta - a

  (2)  Let OB = b

        So, cos (theta - a) = z/b

  (3)  cos a = x/5b

        sin a = y/5b

  (4)  cos (theta - a) = cos theta cos a + sin theta sin a

  (5)  Substitute cos (theta - a) = z/b, cos a = x/5b, sin a = y/5b into

        cos (theta - a) = cos theta cos a + sin theta sin a

                 z/b         = cos theta (x/5b) + sin theta (y/5b)

                 z            = 1/5 [ x cos theta + y sin theta ]

  So, the answer is choice 3

   

  shi fu bai xia zhen lai ler : D

• Forbiddensinner

  28 Jun 09, 13:28

  Originally posted by MasterMoogle:

  shi fu bai xia zhen lai ler : D

  I never said that I am a Maths Genius. If you don't want anymore free Maths Tuition you can go look somewhere else.

  After looking at the workings, I will say that I will have to agree with lecturor Wee's previous statement :

  P.S. Seriously, I feel that these questions are way too convoluted and they defeat the student's self-esteem rather than educate him/her the real concepts. You may wish to give feedback to the setters to provide hints, like dropping perpendiculars.

   

   

• MasterMoogle

  28 Jun 09, 13:29

  Originally posted by Forbiddensinner:

  I never said that I am a Maths Genius. If you don't want anymore free Maths Tuition you can go look somewhere else.

  After looking at the workings, I will say that I will have to agree with lecturor Wee's previous statement :

  P.S. Seriously, I feel that these questions are way too convoluted and they defeat the student's self-esteem rather than educate him/her the real concepts. You may wish to give feedback to the setters to provide hints, like dropping perpendiculars.

   

   

  shi fu dun like that leh, so xiao qi

• qdtimes2

  28 Jun 09, 23:11

  Tip:

  pq = (sin x + sin y)(cos x + cos y) ===> somewhere along the way, you will get sin² x + cos² x which gives you 1

  Try it too :D

   

   

     I still don't get it. after i multipy p and q, i get

    sinx.cosx + sinx.cosy +siny.cosx + siny.cosy = pq

• Forbiddensinner

  29 Jun 09, 18:50

  Originally posted by qdtimes2:

   

     I still don't get it. after i multipy p and q, i get

    sinx.cosx + sinx.cosy +siny.cosx + siny.cosy = pq

  pq = (sin x + sin y)(cos x + cos y) doesn't give you sin² x + cos² x = 1.

  eagle was probably tired and made a mistake there.

  If such a question was to appear as a non-MCQ question, the best method would probably be to reverse-engineer from sin(x+y) to get an appropriate answer.