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Help...I know its easy

Could someone help me solve this equation for C..
.i know its easy but i seem to be getting a possible wrong answer. Im creating a function and i getting a more logical answer when i use another method to solve it but when just solving this equation the answers don't match up with my others. I just a bit skeptical with these things...

Its a practice IA our math teacher has given us


-1.1= 2.68Sin (-1.5 (7 - C)) + 0.28
Reply 1
Try putting everything to one side so you end up with a number = sin(theta). Then you should be able to work out what theta should be. Would you know where to go from there?
Reply 2
1.10.28=2.68sin(1.5(7C))-1.1 - 0.28 = 2.68 \sin(-1.5(7-C))
1.38=2.68sin(1.5(7c))-1.38 = 2.68 sin(-1.5(7-c))
1.382.68=sin(1.5(7c))\frac{-1.38}{2.68} = \sin(-1.5(7-c))

Unparseable latex formula:

\sin^{-1}{\frac{-1.38}{2.68} = -1.5(7-c)



7sin1(1.382.68)1.5=c7 - \frac{\sin^{-1}(\frac{-1.38}{2.68})}{-1.5} = c


Remember to keep calculator in radian mode:
c=6.639
=]

btw, does this forum have latex support?
Reply 3
not how i did it but i got C=-0.011

1) -1.1= 2.68Sin (0.363 (-1.5 - C)) + 0.28
2) -1.1 – 0.28 = 2.68Sin (-0.545 – 0.363C)
3) -1.38 = -1.389 - 2.68Sin(0.363C)
4) -1.38 +1.389 = -0.952C
5) 0.01 = -0.952C
6) 0.01 ÷ -0.952 = C
7) C ≈ -0.011
Reply 4
wait wait wait. line 1:

it's 2.68 sin(-1.5(7-c))....

i believe u did not expand correctly here =]
Reply 5
Its the holidays...why the hell are you working? chill...relax!!!!
Reply 6
pHysiX
-1.1 - 0.28 = 2.68 sin(-1.5(7-C))
-1.38 = 2.68 sin(-1.5(7-c))
[-1.38/2.68] = sin(-1.5(7-c))

sin^-1[-1.38/2.68] = -1.5(7-c)

sin^-1[-1.38/2.68]
7 - ------------------ = c
-1.5

Remember to keep calculator in radian mode:
c=6.639
=]

btw, does this forum have latex support?


http://www.thestudentroom.co.uk/wiki/LaTex

:yep:
Reply 7
hahaha i'm australian. only got 3 more mths until my final exam >.<
Reply 8
Grunx
not how i did it but i got C=-0.011

1) -1.1= 2.68Sin (0.363 (-1.5 - C)) + 0.28
2) -1.1 0.28 = 2.68Sin (-0.545 0.363C)
3) -1.38 = -1.389 - 2.68Sin(0.363C)
4) -1.38 +1.389 = -0.952C
5) 0.01 = -0.952C
6) 0.01 ÷ -0.952 = C
7) C &#8776; -0.011

I'm sturggling to follow this but on line 3 it looks like youve decided to just split the sin(somthing-somthing) to get a number and the sin(C).

You can't do this, it doesn't work. You have to treat everything in the sin() as one entity until the you get rid of the sin()
Reply 9
hmmm if the question is as in the first post, then grunx has already made a mistake in the first line.

however, in terms of methodology, i agree with the muon. if you want to split the sine, use the formula:
sin(A+B)=sin(A)cos(B)+sin(B)cos(A)\sin(A+B) = \sin(A)\cos(B)+\sin(B)\cos(A)
Reply 10
Sorry ive posted the question Wrong,,,

1) -1.1= 2.68Sin (0.363 (-1.5 - C)) + 0.28

Thankyou for your help pHysiX ...Really appreciate youe help
Reply 11
I performed the same calculations as you have physix and came up with the same answer as i previously did even though it was incorrect.
Reply 12
1) -1.1= 2.68Sin (0.363 (-1.5 - C)) + 0.28
2) -1.1 &#8211; 0.28 = 2.68Sin (-0.545 &#8211; 0.363C)
3) -1.38 = -1.389 - 2.68Sin(0.363C)
4) -1.38 +1.389 = -0.952C
5) 0.01 = -0.952C
6) 0.01 ÷ -0.952 = C
7) C &#8776; -0.01


1) -1.1= 2.68Sin (0.363 (-1.5 - C)) + 0.28
2) -1.1 &#8211; 0.28 = 2.68Sin (0.363 (-1.5 - C))
3) -1.38 = 2.68Sin (0.363 (-1.5 - C))
4) -0.515 = Sin (0.363 (-1.5 - C))
5) Sin-1 (-0.515) = (0.363 (-1.5 - C))
6) 1.5- ( (Sin-1 (-0.515))÷(0.363) ) =C
7) C&#8776; -0.01
Reply 13
that's the answer i'm getting as well. u should be right =]
Reply 14
Thanks again mate =)
Grunx
not how i did it but i got C=-0.011

1) -1.1= 2.68Sin (0.363 (-1.5 - C)) + 0.28
2) -1.1 0.28 = 2.68Sin (-0.545 0.363C)
3) -1.38 = -1.389 - 2.68Sin(0.363C)
4) -1.38 +1.389 = -0.952C
5) 0.01 = -0.952C
6) 0.01 ÷ -0.952 = C
7) C &#8776; -0.011


Where has the 0.363 figure come from? I thought it was seven(7) !

The above is not correct.
Reply 16
I typed in the wrong digits and thusly a false answer. Could u help me with the question in my other post titled "Another one of those"...thx Steve

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