LHS = RHS questions (1 Viewer)

[Just.Snaz]

OK, using the identities I mentioned

sinA+sinB = 2cos((A-B)/2)sin((A+B)/2), and
cosA+cosB = 2cos((A-B)/2)cos(A+B)/2))

then the first question is

sin2x+ sin4x + sin6x = 2 cos((4x-2x)/2)sin((2x+4x)/2) +sin6x
= 2 cosxsin3x + 2sin3xcos3x
=2sin3x(cosx + cos3x)
= 2 sin3x(2cos(3x+x)/2)cos(3x-x)/2))
=4sin3xcos2xcosx

The rest of the questions work in a similar fashion.

I don't think it be right telling a year 11 student how to solve these questions using an ext 2 trig identity.. I doubt they'd accept it in an exam..

 

[Slidey]

I don't think it be right telling a year 11 student how to solve these questions using an ext 2 trig identity.. I doubt they'd accept it in an exam..

Incorrect; all methods are accepted in exams.

I highly encourage people to learn and use novel methods, even if they already know how to solve something the 'proper' way; it helps you improve your critical thinking skills, and maybe one day you want know the 'proper' way, but can solve it with another method so you still get marks (it's happened more than once with me).

 

[Just.Snaz]

I had trouble with these questions, some are probably easy but i can't do it.
Any help will be appreciated

These are questions are from 3unit fitzpatrick, exercises 21(c)

49. sin^2(5x) - sin^(3x) = sin8xsin2x.......

49.
RHS = sin8xsin2x
=sin(5x + 3x).sin(5x - 3x)
= [sin5xcos3x + cos5xsin3x][sin5xcos3x - cos5xsin3x]
= sin^2(5x) - sin^2(3x) ---- by expanding difference of two squares

I'll can't get 51. (yet) but i'll try that and the rest later if i can, gotta go to sleep now..

 

[Just.Snaz]

Incorrect; all methods are accepted in exams.

I highly encourage people to learn and use novel methods, even if they already know how to solve something the 'proper' way; it helps you improve your critical thinking skills, and maybe one day you want know the 'proper' way, but can solve it with another method so you still get marks (it's happened more than once with me).

are you 100% positive? cause like.. i dunno.. I always thought you couldn't just state a rule you haven't been taught.. or not meant to have been taught.. my teacher makes this very clear and he marks hsc papers..

but I do know in 4 unit you can use it, even though its been removed from the syllabus.. which is yeah.. contradicting me lol

 

[lolokay]

are you 100% positive? cause like.. i dunno.. I always thought you couldn't just state a rule you haven't been taught.. or not meant to have been taught.. my teacher makes this very clear and he marks hsc papers..

but I do know in 4 unit you can use it, even though its been removed from the syllabus.. which is yeah.. contradicting me lol

the sum/product identities can be easily derived by adding the right angle-sum expansions

eg. sin(A+B) + sin(A-B)
= sinAcosB + sinBcosA + sinAcosB - sinBcosA
= 2sinAcosB

so you just find the correct A and B
eg sin70 + sin10
[A = 40, B = 30]
= 2sin40cos30
= sqrt.3 sin40

I used this type of method to solve the problems posted

 

[Just.Snaz]

the sum/product identities can be easily derived by adding the right angle-sum expansions

eg. sin(A+B) + sin(A-B)
= sinAcosB + sinBcosA + sinAcosB - sinBcosA
= 2sinAcosB

so you just find the correct A and B
eg sin70 + sin10
[A = 40, B = 30]
= 2sin40cos30
= sqrt.3 sin40

I used this type of method to solve the problems posted

oh yeah definitely you can prove it and it'd be cool but I was taught never to use a rule just like that.. BUT if i wanted to, i'd state it, prove it then use it.. I do that with this integral during mechanics.. i think it was 1/(a^2 - x^2) lol can't remember

 

[3unitz]

I'll can't get 51. (yet) but i'll try that and the rest later if i can, gotta go to sleep now..

LHS = 2cos[(85 + 35)/2]cos[(85 - 35)/2] + 2cos[(75 + 45)/2]cos[(75 - 45)/2]

= 2cos60 cos25 + 2cos 60 cos15

= cos 25 + cos 15

= 2 cos[(25 + 15)/2]cos[(25 - 15)/2]

= 2 cos20 cos5

= RHS

 

[Just.Snaz]

LHS = 2cos[(85 + 35)/2]cos[(85 - 35)/2] + 2cos[(75 + 45)/2]cos[(75 - 45)/2]

= 2cos60 cos25 + 2cos 60 cos15

= cos 25 + cos 15

= 2 cos[(25 + 15)/2]cos[(25 - 15)/2]

= 2 cos20 cos5

= RHS

I actually considered that approach but I thought it'd take too long/wouldn't get me anywhere.. I think I need to revise some year 11... and with 5 weeks before trials, that's a bit worrying

 

[shaon0]

Re: 回复: Re: LHS = RHS questions

awesome sig

Thanks...a fourier equation

 

[M@ster P]

Incorrect; all methods are accepted in exams.

I highly encourage people to learn and use novel methods, even if they already know how to solve something the 'proper' way; it helps you improve your critical thinking skills, and maybe one day you want know the 'proper' way, but can solve it with another method so you still get marks (it's happened more than once with me).

Are u sure that u can use methods even if they are not in the syllabus, ie. using products and sums to solve a trig question in a 3unit test paper. I have heard people getting zero marks for doing this.

 

[M@ster P]

LHS = 2cos[(85 + 35)/2]cos[(85 - 35)/2] + 2cos[(75 + 45)/2]cos[(75 - 45)/2]

= 2cos60 cos25 + 2cos 60 cos15

= cos 25 + cos 15

= 2 cos[(25 + 15)/2]cos[(25 - 15)/2]

= 2 cos20 cos5

= RHS

may i ask where did the 2cos60 go?

 

[Iruka]

It got evaluated. Cos60 = 1/2

 

[M@ster P]

oh ok, 2cos60 is basically 1, kool

 

[K@LLo]

hi guys i appreciate your help but i can't still do this question.

61. [sinA + sin(A + B) + sin(A + 2B)] / [cosA + cos(A + B) + cos(A + 2B)] = tan(A+B)

 

[3unitz]

hi guys i appreciate your help but i can't still do this question.

61. [sinA + sin(A + B) + sin(A + 2B)] / [cosA + cos(A + B) + cos(A + 2B)] = tan(A+B)

sinA + sin(A + 2B)
= 2 sin[(A + 2B + A)/2]cos[(A + 2B - A)/2]
= 2 sin (A + B) cos B

.'. sinA + sin(A + B) + sin(A + 2B)
= 2 sin (A + B) cos B + sin(A + B)
= sin (A + B) . (2 cosB + 1) <----- numerator of LHS

cosA + cos(A + 2B)
= 2 cos[(A + 2B + A)/2]cos[(A + 2B - A)/2]
= 2 cos(A + B) cos B

.'. cosA + cos(A + B) + cos(A + 2B)
= 2 cos(A + B) cos B + cos(A + B)
= cos (A + B) . (2 cosB + 1) <----- denominator of LHS

.'. LHS = sin (A + B) . (2 cosB + 1) / cos (A + B) . (2 cosB + 1)
= sin (A + B) / cos (A + B)
= tan (A + B)
= RHS

 

[K@LLo]

wow thx 3unitz, did the question take you long to figure out?, because your working out was very logical.

 

[lolokay]

hi guys i appreciate your help but i can't still do this question.

61. [sinA + sin(A + B) + sin(A + 2B)] / [cosA + cos(A + B) + cos(A + 2B)] = tan(A+B)

as I said earlier, let C = A + B, then get rid of the A's

[top]sin(C - B) + sinC + sin(C + B) = sinC(2cosB + 1)
[bottom]cos(C-B) + cosC + cos(C + b) = cosC(2cosB + 1)
top/bottom = sinC/cosC = tanC
= tan(A+B)