trigo identities

• wishboy

  13 Sep 07, 20:40

  got a few qns i cannot solve or solved halfway
  
  Prove the following identity.
  ( 3 - 6 cos² x ) / ( sin x - cos x ) = 3 ( sin x + cos x )
  
  Prove that
  
  (a) cot x + tan x = cosec x sec x
  (b) cosec² x + sec² x = cosec² x sec² x
  
  Hence, deduce that (cot x + tan x )² = cosec² x + sec² x
  
  5. Find all the angles between 0° and 360° which satisfy the following equations.
  
  (d) cos ( 2x - 40° ) = 0.8
  this qn i found 2 answers which are 38.4° and 218.4°, but there are 2 more answers which are 1.6° and 181.6°.
  i dunno how to find those 2.
  
  (e) cot ( 2x + 10° ) = - 0.5
  this qn i get the answers all wrong de
  my 1st step was
  tan ( 2x + 10° ) = - 2
  answers: 53.3°, 143.3°, 233.3°, 323.3°
  
  will appreciate any help
• eagle

  13 Sep 07, 21:06

  Originally posted by wishboy:

  Prove the following identity.
  ( 3 - 6 cos² x ) / ( sin x - cos x ) = 3 ( sin x + cos x )

  ( 3 - 6 cos² x ) = 3 - 3cos² x -3cos² x = 3( sin² x - cos² x )
  
  = 3 (sin x + cos x) (sin x - cos x)
  
  after divide u will get answer
  
  

  Originally posted by wishboy:

  Prove that
  
  (a) cot x + tan x = cosec x sec x
  (b) cosec² x + sec² x = cosec² x sec² x
  
  Hence, deduce that (cot x + tan x )² = cosec² x + sec² x

  a) cos x / sin x + sin x / cos x = ( cos²x + sin²x ) / (sin x cos x)
  = 1 / (sin x cos x) = cosec x sec x
  
  b) 1 / sin² x + 1 / cos² x = ( cos² x + sin² x ) / (sin² x cos² x)
  = 1 / (sin² x cos² x) = cosec² x sec² x
  
  (part a)² = (cot x + tan x )² = cosec² x sec² x = cosec² x + sec² x (deduced)
  
• ^tamago^

  13 Sep 07, 21:06

  Prove: (3-6 cos²x)/(sin x-cos x) = 3(sin x+cos x)
  
  LHS = (3-6 cos²x)/(sin x-cos x)
  = 3(1-2cos²x)/(sin x-cos x)
  = 3(sin²x-cos²x)/(sin x-cos x)
  = 3(sin x+cos x)
  = RHS (proven)
• eagle

  13 Sep 07, 21:09

  Originally posted by wishboy:

  5. Find all the angles between 0° and 360° which satisfy the following equations.
  
  (d) cos ( 2x - 40° ) = 0.8

  U must know
  0 < x < 360
  0 < 2x < 720
  -40 < 2x - 40 < 680
  
  hence, 2x - 40 = -36.87, 36.87, 323.13, 396.87
  x = 1.6, 38.4, 181.6, 218.4
  
  understand part d, then try part e by yourself
  
• ^tamago^

  13 Sep 07, 21:14

  cos(2x-40°) = 0.8
  For -40°<2x<680°,
  2x-40°=-36.86°, 36.86°, 360-36.86°, 360+36.86° = -36.86°, 36.86°, 323.14°, 396.86°
  2x=3.14°, 76.86°, 363.14°, 436.86°
  x=1.57°, 38.43°, 181.57°, 218.43° = 1.6°, 38.4°, 181.6°, 218.4°
  
• ^tamago^

  13 Sep 07, 21:21

  cot(2x+10°) = -0.5
  tan(2x+10°) = -2
  For 10°<2x<730°,
  2x+10°= 180-63.43°,360-63.43°, 540-63.43°, 720-63.43° = 116.57°, 296.56°, 476.57°, 656.56°
  2x= 106.57°, 286.56°, 466.57°, 646.56°
  x= 53.29°, 143.28°, 233.29°, 323.28° = 53.3°, 143.3°, 233.3°, 323.3°
  
• wishboy

  13 Sep 07, 23:21

  thanks for the replies!
  lol ok i peep liao
• ^tamago^

  13 Sep 07, 23:37

  Originally posted by wishboy:

  thanks for the replies!
  lol ok i peep liao

  ok. cos eagle say do urself so i dun.....