LHS = RHS questions (1 Viewer)

[K@LLo]

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)

47. sin2x + sin4x + sin6x = 4cosxcos2xsin3x

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

51. cos35 + cos45 + cos75+ cos85 = 2cos5cos20

54. sin10 + cos40 = sin70

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

Thx

 

[shaon0]

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)

47. sin2x + sin4x + sin6x = 4cosxcos2xsin3x

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

51. cos35 + cos45 + cos75+ cos85 = 2cos5cos20

54. sin10 + cos40 = sin70

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

Thx

47. Use double-angle formulae to break the equation down in LHS and make it equal RHS.
49. Let LHS= sin5x.sin5x -sin3x.sin3x
Break them into their components.
51. break them into their components....EASY
54. Same as above.....EASY
61. Break them up into their coomponents....EASY
LHS will end up being:
sin (a+B) / cos (A+B)

 

[tommykins]

回复: Re: LHS = RHS questions

Rofl you say break it up into their components, how about try it before you actually say it's easy?

 

[K@LLo]

are u talking to me tommykins?

 

[K@LLo]

47. Use double-angle formulae to break the equation down in LHS and make it equal RHS.
49. Let LHS= sin5x.sin5x -sin3x.sin3x
Break them into their components.
51. break them into their components....EASY
54. Same as above.....EASY
61. Break them up into their coomponents....EASY
LHS will end up being:
sin (a+B) / cos (A+B)

thx i will use ur advice

 

[K@LLo]

sorry but i already tried out ur advice (just realised) and i could not do the questions so yeah basically im asking for the solutions to each question if thats not too much

 

[tommykins]

回复: Re: LHS = RHS questions

are u talking to me tommykins?

No, I'm talking to shan0. The reason being is, notice the sin/cos[ax] where a > 3 makes it absolutely ridiculous, as expanding this will take ages (unless you do it the extension 2 way, which you've obviously have no learnt).

I'm sorry I can't help you however, I haven't touched math all holidays and my mathematical thinking is up shit creek.

 

[K@LLo]

which question is this sorry?

 

[K@LLo]

Re: 回复: Re: LHS = RHS questions

No, I'm talking to shan0. The reason being is, notice the sin/cos[ax] where a > 3 makes it absolutely ridiculous, as expanding this will take ages (unless you do it the extension 2 way, which you've obviously have no learnt).

I'm sorry I can't help you however, I haven't touched math all holidays and my mathematical thinking is up shit creek.

oh if u meant that there is more then x is more then 3, there is no need to worry i have learnt the products to sums and sums to differences.

 

[u-borat]

Re: 回复: Re: LHS = RHS questions

No, I'm talking to shan0. The reason being is, notice the sin/cos[ax] where a > 3 makes it absolutely ridiculous, as expanding this will take ages (unless you do it the extension 2 way, which you've obviously have no learnt).

I'm sorry I can't help you however, I haven't touched math all holidays and my mathematical thinking is up shit creek.

wats the extension 2 way?

fuck, you're scaring the shit out of me?

 

[shaon0]

Re: 回复: Re: LHS = RHS questions

wats the extension 2 way?

fuck, you're scaring the shit out of me?

Something to do with imaginary number.

 

[shaon0]

Re: 回复: Re: LHS = RHS questions

Rofl you say break it up into their components, how about try it before you actually say it's easy?

lol....it is easy...but it takes a while.
I have no time since i have to do assignments.

 

[Iruka]

He is talking about using deMoivre's theorem (and Pascal's triangle).

Most of these questions can be solved comparatively quickly if you use the identity

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

and related identities.

There is a derivation part way down this page:

http://www.acts.tinet.ie/sumstoproductsproducts_673.html

 

[K@LLo]

ok i know how to do "products as sums and differences" and "sums and differences to products" just these questions are weird and i cant do them

 

[tommykins]

回复: Re: 回复: Re: LHS = RHS questions

wats the extension 2 way?

fuck, you're scaring the shit out of me?

DMT's theorem with sinax being Im(z^a) and cosax being Re(z^a) and z = cosx + isinx

@ K@LLo
can you put the identities here? i'll see what i can do with it

 

[K@LLo]

ok

products to sum
cos A cos B = ½ cos(A−B) + ½ cos(A+B) sin A sin B = ½ cos(A−B) − ½ cos(A+B)
sin A cos B = ½ sin(A+B) + ½ sin(A−B)
cos A sin B = ½ sin(A+B) − ½ sin(A−B)

sums to products

cos u + cos v = 2 cos(½(u+v)) cos(½(u−v)) cos u − cos v = −2 sin(½(u+v)) sin(½(u−v))
sin u + sin v = 2 sin(½(u+v)) cos(½(u−v))
sin u − sin v = 2 sin(½(u−v)) cos(½(u+v))

 

[Iruka]

ok i know how to do "products as sums and differences" and "sums and differences to products" just these questions are weird and i cant do them

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.

 

[K@LLo]

other solutions for other questions too pretty please

 

[lolokay]

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)

47. sin2x + sin4x + sin6x = 4cosxcos2xsin3x

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

51. cos35 + cos45 + cos75+ cos85 = 2cos5cos20

54. sin10 + cos40 = sin70

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

Thx

49. is a difference of two squares, then converting 2sinXcosX into sin2X
51. Solve each pair normally, then sub in exact value of cos60
54. write cos40 as sin50. solve normally, then sub in exact value of sin30
61. let A + B = C, then get rid of the A's

 

[u-borat]

Re: 回复: Re: 回复: Re: LHS = RHS questions

DMT's theorem with sinax being Im(z^a) and cosax being Re(z^a) and z = cosx + isinx

@ K@LLo
can you put the identities here? i'll see what i can do with it

ah right, scared the absolute shit out of me. thanks for clarifying.