Finding General Solution (1 Viewer)

hsc2013

New Member
Joined
Jan 14, 2012
Messages
20
Gender
Male
HSC
2013
I was doing my trig homework and i have some difficulties with these questions. Could anyone please show me the full working out to these questions so i could see where have i gone wrong.
For questions 8, 10 and 48 my solutions was way off the correct answer and for question 9, I got my solutions in terms of sin but the answer for some reason is in terms of cos. Could you guys tell me is my answer wrong or is the homework answer wrong

8. cosx/2 =1
9. 2 sin^2x =1
10. 4cos^2x-1 =0
48. sin2x =tanx

I also want to ask is there a formula for sin3x, cos3x and tan 3x and is it the same as sin 4x, sin 5x and so on.

Thank you
 

812

New Member
Joined
Feb 25, 2011
Messages
28
Location
The gates of Mt Olympus
Gender
Female
HSC
2012
Yes there are triple angle formulas but you are not expected to remember them. However, questions can be asked for you to derive triple angle formulas.


Now, I'm assuming these questions are only asking for solutions 0 < x < 360..

Question 8.
cos(x/2) = 1
cos0 = 1,
therefore x/2 = 0,
x = 0

Question 9.
2sin^2(x) = 1,
using double angle formula ---> (-cos2x) + 1 = 1
--> cos2x = 0,
since there is a double angle, you need to have 4 solutions: cos90 = 0, cos270 = 0, cos(90+360) = 0, cos(270+360) = 0,
and therefore 2x = 90 or 270 or 450 or 630
x = 45, 135, 225, 315

Question 10.
4cos^2(x) - 1 = 0,
using double angle formula --> 2cos2x = 0
--> cos2x = 0......
the rest is the same as question 9

Question 48.
sin2x = tanx,
2sinxcosx = sinx/cosx
2sinxcos^2(x) - sinx = 0
2sinx(1 - sin^2(x)) - sinx = 0
2sinx - 2sin^3(x) - sinx = 0
sinx - 2sin^3(x) = 0
1 - 2sin^2(x) = 0
cos2x = 0
the rest is the same as question 9
 
Last edited:

RealiseNothing

what is that?It is Cowpea
Joined
Jul 10, 2011
Messages
4,591
Location
Sydney
Gender
Male
HSC
2013
Sin3x= 3sinx - 4(sin^3)x

Cos3x = 4(cos^3)x -3cosx

Tanx = [3tanx - (tan^3)x] / [1 - 3(tan^2)x]

As 812 said, you don't have to memorise them but you may be asked to derive them. All you have to do to derive them is change sin3x to sin(2x+x) and go from there.
 

hsc2013

New Member
Joined
Jan 14, 2012
Messages
20
Gender
Male
HSC
2013
Sorry guys, I have a few more homework questions I couldn't do. I attempted the but my answers are wrong:vcross: Could you guys provide me with the working out and solutions as well.

Prove the following

41. sin5x-sin3x
_________ = tanx
cos5x + cos 3x


42. sinx + siny tanx+y
_________ = ______
cosx + cosy 2


47. sin5x +sin3x -2sin2xcosx = 2sin2xcos3x
48. sin2x +sin4x + sin6x = 4cosx cos2x sin3x
50. sin 35 -sin25 = root3 sin5

Please note that question 41 and 42 are fractions and that all questions are required to prove LHS=RHS

Thanks
 
Last edited:

812

New Member
Joined
Feb 25, 2011
Messages
28
Location
The gates of Mt Olympus
Gender
Female
HSC
2012
Bro you're asking a lot of questions on the same stuff.

I think it's better if you go back and practise more basic questions and consolidate your foundation. There is no point in us giving you the solutions if you don't understand the concepts.
 

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

Top