A maths - trigo

• bonkysleuth

  9 Dec 08, 13:26

  i can't solve this despite numerous attempts. please help.

   

  If 270deg < x < 360deg, simplify [sq.rt2 + (sq rt. 2 + 2 cos x)]

   

  Answer's 2 sin x/4 but i got 2 cos x/4.

   

  And next, may i know what happens when you have sqrt (4 cos^2 0.5x)? can it be reduced to 2 cos 0.5x?

• eagle

  9 Dec 08, 13:56

  first part, can you rephrase more clearly?

  use things like sqrt(2). You varied between sq.rt2 and sq rt. 2, etc etc... quite unclear.

   

  And next, may i know what happens when you have sqrt (4 cos^2 0.5x)? can it be reduced to 2 cos 0.5x?

  Yes, you are right.

• bonkysleuth

  10 Dec 08, 00:19

  Originally posted by eagle:

  first part, can you rephrase more clearly?

  use things like sqrt(2). You varied between sq.rt2 and sq rt. 2, etc etc... quite unclear.

   

   

  
  the sq.rt 2 in the box " [ ....] is the first one while the one with "(.....)" consists of the 2nd square root.

• eagle

  10 Dec 08, 08:36

  Originally posted by bonkysleuth:

  
  the sq.rt 2 in the box " [ ....] is the first one while the one with "(.....)" consists of the 2nd square root.

  This?

  

  
  

  or

  
  

  
• bonkysleuth

  10 Dec 08, 14:03

  Originally posted by eagle:

  This?

  

  
  

  or

  
  

  

  
  the second one . :)

• eagle

  10 Dec 08, 14:11

  I'm rather busy right now, so won't do yet, but the answer of 2 sin (x/4) is correct.

  Hopefully someone else can help.

   

  Hint: To check correctness (in exams as well), sub in a value. I used x = 300 degrees in my quick check.

   

• Mikethm

  10 Dec 08, 19:01

  
  

  

  Removed due to misread. :P

• eagle

  10 Dec 08, 19:12

  Originally posted by Mikethm:

  = sqrt{ 2 + sqrt [ 2 + 2(2cos^2 0.5x -1)]}

  = sqrt { 2 + sqrt [4cos^2 0.5x]}

  = sqrt ( 2 + 2cos^2 0.5x )

  = sqrt [ 2 + 2(2cos^ 0.25x -1)]

  = sqrt ( 4cos^2 0.25x)

  = 2cos 0.25x

  hmmm... rechecking my workings...

  sqrt [4cos^2 0.5x] = -2 cos 0.5 x

  because sqrt of anything is a positive number
  since 135deg < 0.5x < 180deg

  Hence, cos 0.5 x is negative
  Thus you need to put a negative sign in front to turn it positive.

  should be able to solve it after the negative sign

• bonkysleuth

  10 Dec 08, 21:59

  Originally posted by eagle:

  sqrt [4cos^2 0.5x] = -2 cos 0.5 x

  because sqrt of anything is a positive number
  since 135deg < 0.5x < 180deg

  Hence, cos 0.5 x is negative
  Thus you need to put a negative sign in front to turn it positive.

  should be able to solve it after the negative sign

  
  I still do not quite understand. I tried using the method you prescribed but somehow, the solution turns out more confusing and I kept getting 2 cos 1/4 x with my previous method. If possible, can you explain it again? thanks.

• eagle

  11 Dec 08, 00:03

  
  
  

  
  

  Since 270 < x < 360, 135 < 0.5x < 180

  Thus, cos 0.5x is negative

  However, since the square root of any number is always positive, hence,

  so that it is positive.... ===> Not 100% sure if this logic is correct......

  
  

  But if we follow from the top,

  
  
  

  

  
  

  Since 67.5 < 0.25x < 90, 0.25x is in the first quadrant, hence sine is positive

  
  

  Thus,

  

  
• Mikethm

  11 Dec 08, 01:34

  Originally posted by eagle:

  sqrt [4cos^2 0.5x] = -2 cos 0.5 x

  because sqrt of anything is a positive number
  since 135deg < 0.5x < 180deg

  Hence, cos 0.5 x is negative
  Thus you need to put a negative sign in front to turn it positive.

  should be able to solve it after the negative sign

  oh heck i didn't notice that a range for x was given coz I read the question from the image u posted haha... my bad.

• eagle

  11 Dec 08, 08:14

  Originally posted by Mikethm:

  oh heck i didn't notice that a range for x was given coz I read the question from the image u posted haha... my bad.

  I realised that after I saw you posting that you tested it with 100 degrees :D