A.Maths - Further Trigo

• wishboy

  28 Oct 07, 19:00

  5. Find all angles between 0° and 360° which satisfy the following equations.
  
  3 sin x = 2 cos (x + 45°)
  
  my ans different from answer key
  correct ans is 17.8°, 197.8°.
• eagle

  28 Oct 07, 19:03

  Originally posted by wishboy:

  5. Find all angles between 0° and 360° which satisfy the following equations.
  
  3 sin x = 2 cos (x + 45°)
  
  my ans different from answer key
  correct ans is 17.8°, 197.8°.

  how u do?
  o levels or a levels?
  
  got learn how to expand cos(x + 45°) a not before?
  
  edit: I just realised topic says A maths. I think of an o level method that get back to u
• wishboy

  28 Oct 07, 19:07

  ya o lvl method
  i noe that cos (a+b) = cos a cos b - sin a sin b
• eagle

  28 Oct 07, 19:16

  Originally posted by wishboy:

  ya o lvl method
  i noe that cos (a+b) = cos a cos b - sin a sin b

  if u know the expansion, you can easily expand it
  
  then u make one side all sin x, one side all cos x
  
  then u divide the cos x over to get tan x, and divide the left side number to the right side.
  
  u should then get
  
  tan x = sqrt(2) / { 3 + sqrt(2) }
  
  which you should get x = 17.8 degrees. Add 180 degrees for the other answer becuz this is tangent.
  
  But I'm not sure if this expansion is in o level syllabus
• wishboy

  28 Oct 07, 19:29

  (3 sin x) / 2 = 3/2 sin x
  
  EDIT:
  
  and since cos 45° = sin 45°,
  cos x cos 45° - sin x sin 45°
  = sin 45° (cos x - sin x)
  
  are they correct?
• eagle

  28 Oct 07, 19:32

  Originally posted by wishboy:

  (3 sin x) / 2 = 3/2 sin x
  correct?

  correct
• wishboy

  28 Oct 07, 19:39

  err the 2nd one?
  i edited the post
• ^tamago^

  28 Oct 07, 19:39

  Originally posted by wishboy:

  5. Find all angles between 0° and 360° which satisfy the following equations.
  
  3 sin x = 2 cos (x + 45°)

  best to do by formulae to avoid doubt
  
  3 sin x = 2 cos (x + 45°)
  3 sin x = 2 cos x cos 45° - 2 sin x sin 45°
  3 sin x = sqrt(2)(cos x-sin x)
  
  
  
  
  
  x = 17.76°, 180+17.76° (see ASTC)
  = 17.8°, 197.8°
• eagle

  28 Oct 07, 19:40

  Originally posted by wishboy:

  err the 2nd one?
  i edited the post

  correct
  
  anyway just use the value of sqrt(2) / 2 for sin 45
• ^tamago^

  28 Oct 07, 19:40

  Originally posted by eagle:

  if u know the expansion, you can easily expand it
  
  then u make one side all sin x, one side all cos x
  
  then u divide the cos x over to get tan x, and divide the left side number to the right side.
  
  But I'm not sure if this expansion is in o level syllabus

  yeah it is.
• wishboy

  28 Oct 07, 19:42

  lol somehow i got the answer, although the methods i used look so much alike
  thx for ur help
• wishboy

  28 Oct 07, 19:43

  Originally posted by eagle:

  correct
  
  anyway just use the value of sqrt(2) / 2 for sin 45

  haven learn
• ^tamago^

  28 Oct 07, 19:53

  Originally posted by wishboy:

  haven learn

  it's in the syllabus and in the textbook. try to memorise this table. it's helpful all the way to uni.
  
  
• ^tamago^

  28 Oct 07, 20:03

  for tan,
  
  tan 30 is sqrt(3)/3
  tan 45 is 1
  tan 60 is sqrt(3)
  
  also useful.
• wishboy

  28 Oct 07, 20:08

  ooo so i learn liao... but forgot
  for my textbook they give sin 45° = 1/sqrt(2)
  value same so nvm
  
  another qn,
  
  0° < x < 360°
  
  tan 5x = tan 2x
  
  find all values of x
  
  this qn got * meaning more difficult than normal ones
• eagle

  28 Oct 07, 20:11

  i always teach my students this method
  
  sin 0 = sqrt(0)/2
  sin 30 = sqrt(1)/2
  sin 45 = sqrt(2)/2
  sin 60 = sqrt(3)/2
  sin 90 = sqrt(4)/2
  
  cos just reverse everything
  
  but hor... everything is already second nature to me
• ^tamago^

  28 Oct 07, 21:30

  
  
  
• eagle

  28 Oct 07, 21:35

  Originally posted by ^tamago^:

  

  First 2 line RHS got problem
  
  and don't need do till so mah fan mah
  
  tan 5x = tan 2x
  0 < 5x < 1800
  0 < 2x < 720
  
  0 < 5x - 2x < 1080
  
  then
  
  5x = 2x + 180, 2x + 360, 2x + 540, 2x + 720, 2x + 900
  3x = 180, 360, 540, 720, 900
  x = 60, 120, 180, 240, 300
• ^tamago^

  28 Oct 07, 21:41

  Originally posted by eagle:

  First 2 line RHS got problem
  
  and don't need do till so mah fan mah

  changed.
  
  how come
  tan 5x = tan 2x
  can jump
  5x = 2x
  
  
• wishboy

  28 Oct 07, 21:47

  Originally posted by ^tamago^:
  [/quote]
  thanks
  
  [quote]Originally posted by eagle:
  First 2 line RHS got problem
  
  and don't need do till so mah fan mah
  
  tan 5x = tan 2x
  0 < 5x < 1800
  0 < 2x < 720
  
  0 < 5x - 2x < 1080
  
  then
  

  5x = 2x + 180, 2x + 360, 2x + 540, 2x + 720, 2x + 900
  3x = 180, 360, 540, 720, 900
  x = 60, 120, 180, 240, 300

  how u get to that bolded step?
  
• eagle

  28 Oct 07, 21:51

  Originally posted by ^tamago^:

  changed.
  
  how come
  tan 5x = tan 2x
  can jump
  5x = 2x
  
  

  okok I skipped a step
  
  tan 5x = tan 2x
  
  5x = arctan (tan 2x) +180, arctan (tan 2x) + 360, and so on.......
• eagle

  28 Oct 07, 21:52

  Originally posted by wishboy:

  how u get to that bolded step?
  

  see the above post if it answers your question
• ^tamago^

  28 Oct 07, 21:59

  Originally posted by eagle:

  okok I skipped a step
  
  tan 5x = tan 2x
  
  5x = arctan (tan 2x) +180, arctan (tan 2x) + 360, and so on.......

  uncle add maths dun have leh. unless i didn't read this formulae.
• wishboy

  28 Oct 07, 22:02

  wth is arctan?
  not in my syllabus
  nvm no need to answer this
• HyuugaNeji

  28 Oct 07, 22:02

  there isnt a need for a formula. This is by observation that tan x repeats itself every 180 degree while sin x and cos x repeats itself every 360 degrees.