Trigonometry monster equations! Help

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rexexdesign

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Feb 4, 2008
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6
Hello Everyone,

thanks already for reading, and if you can help me out, even more thanks! (Handshake)

My teacher surprised us today with extra credit homework. He said we can ask anyone for help. He thinks that not too many people will be able to solve his problems.

I will list them all and number them. If any one of these equations look like you can do it, feel free to post the answer and the corresponding number, so that questions won't be answered twice.

Thanks so much!

1) cos7x * cos13x = cosx * cos19x
2)sin2x * cos8x + sin6x = 0
3)cos9x - cos7x + cos3x - cosx = 0
4)2sin^2(x) - cos2x + sin2x = 0
5)2cosx + 3sinx = 1
6)4sin^4(x) -1 = 5 cos^2(x)
7)sinx + sin^2(x) + cos^3(x) = 0
 
I think you misunderstand the purpose of this site. We, well most of us, are here to help those with their math problems, not do them for you.

How about you provide what you have tried and we'll help you point out any errors & guide you in the right direction.
 
rexexdesign said:
1) cos7x * cos13x = cosx * cos19x
2)sin2x * cos8x + sin6x = 0
3)cos9x - cos7x + cos3x - cosx = 0
4)2sin^2(x) - cos2x + sin2x = 0
5)2cosx + 3sinx = 1
6)4sin^4(x) -1 = 5 cos^2(x)
7)sinx + sin^2(x) + cos^3(x) = 0
Click to expand...
There are sites at which you can pay to have these done for you.
Of course that is cheating; but the folks at those sites don’t seem to care.
That is not true of this site.
Please go elsewhere if all you want are answers
 
rexexdesign said:
Hello Everyone,

thanks already for reading, and if you can help me out, even more thanks! (Handshake)

My teacher surprised us today with extra credit homework. He said we can ask anyone for help. He thinks that not too many people will be able to solve his problems.

I will list them all and number them. If any one of these equations look like you can do it, feel free to post the answer and the corresponding number, so that questions won't be answered twice.

Thanks so much!

1) cos7x * cos13x = cosx * cos19x
2)sin2x * cos8x + sin6x = 0
3)cos9x - cos7x + cos3x - cosx = 0
4)2sin^2(x) - cos2x + sin2x = 0
5)2cosx + 3sinx = 1
6)4sin^4(x) -1 = 5 cos^2(x)
7)sinx + sin^2(x) + cos^3(x) = 0
Click to expand...
dUPLICATE POST

http://www.mathhelpforum.com/math-help/ ... -help.html
 
Ok, I will give it a shot. I just gave up once he pretty much said that these are not common equations and that we can ask anyone (meaning we will need the help). Let me try some problems and post if I am stuck, thanks!
 
rexexdesign said:
Hello Everyone,

1) cos7x * cos13x = cosx * cos19x
Use
cos A*cosB = 1/2[cos(A+B) + cos(A-B)]

2)sin2x * cos8x + sin6x = 0
use
sinA*cosB = 1/2[sin(A+B) + sin(A-B)]

3)cos9x - cos7x + cos3x - cosx = 0

[cos9x - cosx] - [cos7x - cos3x]=0

[cos(5+4)x - cos(5-4)x] - [cos(5+2)x - cos(5-x)] = 0

4)2sin^2(x) - cos2x + sin2x = 0
5)2cosx + 3sinx = 1
6)4sin^4(x) -1 = 5 cos^2(x)
7)sinx + sin^2(x) + cos^3(x) = 0
Click to expand...
 
As you said, I used the formula you gave. This is what I got

=1/2[sin(2x+8x)+sin(2x-8x)]+sin6x=0
=1/2 [sin10x-sin6x]=0
=1/2*sin10x-1/2*sin6x-sin6x=0

I am wondering now, if this equation is any easier (can I say -1/2sin6x-sin6x = -3/2sin6x ?)


2)sin2x * cos8x + sin6x = 0
use
sinA*cosB = 1/2[sin(A+B) + sin(A-B)]
 
rexexdesign said:
As you said, I used the formula you gave. This is what I got

1/2[sin(2x+8x)+sin(2x-8x)] + sin6x=0

1/2 [ sin10x - sin 6x] + sin 6x = 0

1/2[sin 10x + sin 6x] = 0

sin 10x + sin 6x = 0

sin(8x + 2x) + sin (8 -2) x = 0

sin8x * cos 2x + cos 8x * sin 2x + sin 8x * cos 8x - sin 2x * cos 8x = 0

2 * sin 8x * cos 2x = 0

Then either

sin 8x = 0 >>>> 8x = n * (pi) >>> x = n/8 * (pi)

or

cos 2x = 0 >>>> 2x = (n + 1/2) * (pi) >>>> x = (n + 1/2) *(pi)/2

............edited....
Click to expand...
 
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