• Want to help us with this year's BoS Trials?
    Let us know before 30 June. See this thread for details
  • Looking for HSC notes and resources?
    Check out our Notes & Resources page

Examine's 2u horrors (1 Viewer)

ravana94

New Member
Joined
Feb 18, 2012
Messages
7
Gender
Male
HSC
2012
Hey guys I was wondering if you couldo this trig differentiation( because i couldn't :p)

11a) show that 1/2(sin(m+n)x+sin(m-n)x) = sinmx cosnx. (this is the easy bit)
b) Hence, without using the product rule, find the derivative of sin mx cos nx ( :p)
 

bleakarcher

Active Member
Joined
Jul 8, 2011
Messages
1,509
Gender
Male
HSC
2013
Just utilise part a). So:
d/dx[sin(mx)cos(nx)]=d/dx[(1/2)[sin(m+n)x+sin(m-n)x]=(1/2)[(m+n)cos(m+n)x+(m-n)cos(m-n)x]
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Hey guys I was wondering if you couldo this trig differentiation( because i couldn't :p)

11a) show that 1/2(sin(m+n)x+sin(m-n)x) = sinmx cosnx. (this is the easy bit)
b) Hence, without using the product rule, find the derivative of sin mx cos nx ( :p)
 

ravana94

New Member
Joined
Feb 18, 2012
Messages
7
Gender
Male
HSC
2012
Hey thanks for your working out it really helped, time to do some more math :p
 

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
The question came up in Terry Lee integrals that trigonometric expansion.

Quite fun.
 

Timske

Sequential
Joined
Nov 23, 2011
Messages
794
Gender
Male
HSC
2012
Uni Grad
2016
trolollolol lalalala lalalal trolololol lala hooo haha hooo haha
 

Examine

same
Joined
Dec 14, 2011
Messages
2,377
Gender
Undisclosed
HSC
2013
Need help with this question.

The roots of the quadratic equation
x^2-(2k+4)x+(k^2+3k+2)=0
are non-zero and one of the roots is twice the other root, calculate the value of k.
 

Carrotsticks

Retired
Joined
Jun 29, 2009
Messages
9,494
Gender
Undisclosed
HSC
N/A
Need help with this question.

The roots of the quadratic equation
x^2-(2k+4)x+(k^2+3k+2)=0
are non-zero and one of the roots is twice the other root, calculate the value of k.
Let the roots be a and 2a (since one is twice the other).

Sum of roots is 2k+4, which is equal to 3a (I added the two roots).

Thus, a = 2/3 * (k+2)

P(a) = 0 since a is a root.

Solve for k.
 

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
Note: Harrians in this thread. It is important you attend all classes. Also listen and do the homework.
 

Examine

same
Joined
Dec 14, 2011
Messages
2,377
Gender
Undisclosed
HSC
2013
Let the roots be a and 2a (since one is twice the other).

Sum of roots is 2k+4, which is equal to 3a (I added the two roots).

Thus, a = 2/3 * (k+2)

P(a) = 0 since a is a root.

Solve for k.
Can you clarify this step?
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top