Additional Maths: Trigonometric Fnc

• Nathpoop

  29 Aug 09, 18:03

  How do i solve the following questions?

  Find the values of θ, where θ is more than or equal to 0deg and less than or equal to 360deg

  1(a)

  (sin θ - 1)(sin θ +1) = 0

  Ans: 90deg, 270deg

  (b)

  sin θ (2 cos θ - 3) = 0

  Ans: 0deg, 180deg, 360deg.

  Thanks in advance! (:

• Mikethm

  29 Aug 09, 18:29

  a)

  sin^2 x -1 = 0

  sin^2 x = 1

  sin x = +/-1

  Reference angle = 90

  x = 90 or 270.

   

  b)

  sin x(2cos x  -3) = 0

  sin x = 0 or cos x = 3/2 (NA)

  Reference angle = 0

  x = 0, 180 or 360.

  Sec 3 not Sec 4 right?

• SBS n SMRT

  29 Aug 09, 19:40

  Yes, trigo is in sec 3

• jayh272416

  29 Aug 09, 19:42

  Originally posted by Mikethm:

  a)

  sin^2 x -1 = 0

  sin^2 x = 1

  sin x = +/-1

  Reference angle = 90

  x = 90 or 270.

  you can simply jump to sin x = +/-1 lol. since its already statede (sin x +1) (sin x -1) = 0 lol

• SBS n SMRT

  29 Aug 09, 19:46

  Originally posted by jayh272416:

  you can simply jump to sin x = +/-1 lol. since its already statede (sin x +1) (sin x -1) = 0 lol

  no lah

  (sinx+1)(sinx-1)=o

  sinx+1 = 0 or sinx-1=0

  sinx=+/- 1

  this way of presenting is clearer

• wee_ws

  29 Aug 09, 20:17

  Hi Nathpoop,

    Schools like to set questions like

    (sin θ - 1)(sin θ + 1) = -1/2,

    so be sure to know how to tackle these too. Thanks!

  Cheers,
  Wen Shih

• Nathpoop

  29 Aug 09, 23:46

  I don't get it why is the answer for part(b) 0deg, 180deg, 360deg. Like the ASTC which quadrant they are in? And how do i arrive to the answer from the reference angle, 0 deg?

• Mikethm

  30 Aug 09, 00:27

  Originally posted by jayh272416:

  you can simply jump to sin x = +/-1 lol. since its already statede (sin x +1) (sin x -1) = 0 lol

  Thank you but my way is less time consuming than writing "sinx =1 or sinx = -1" and clearer than jumping straight to "sinx =+/-1". Lumping it together allow me to do with one SATC diagram instead of 2.

  To Nathpoop:

  Follow the arrows in calculating.

  

   

• Nathpoop

  30 Aug 09, 01:07

  Total there are for quads

  0deg in  1st quad

  180 in 2nd quad

  360 in 4th quad

  am i correct?

  If i refer to your diagram wouldn't it be 4 answers instead of 3?

  -

  x = 0, 180 - 0, 180 + 0, 360 - 0

     = 0, 180, 180, 360

  therefore x = 0, 180, 360.

  Am i right to show this as my working?

• Mikethm

  30 Aug 09, 02:16

  Originally posted by Nathpoop:

  Total there are for quads

  0deg in  1st quad

  180 in 2nd quad

  360 in 4th quad

  am i correct?

  If i refer to your diagram wouldn't it be 4 answers instead of 3?

  -

  x = 0, 180 - 0, 180 + 0, 360 - 0

     = 0, 180, 180, 360

  therefore x = 0, 180, 360.

  Am i right to show this as my working?

  Depend on what you would show for say... (x-2)(x-2)=0

  Personally I would just show "x=2" instead of "x=2 or x=2"

  Nothing wrong with showing repeated roots but would annoy some. So I wouldn't.