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Simultaneous trigonometric equations

  • Thread starter Ry122
  • Start date
565
2
0=-85.7cos(theta)+bcos30-85.7
0=-85.7sin(theta)+bsin30
Can someone please tell me what I am doing wrong.
theta=sin^-1(bsin30/85.7)
then sub that into the first equation.
I then solved the equation on my calculator and it gave me 9999999999 which i know is wrong.
 
458
0
Are you trying to solve the equation for b or theta? It's hard to tell.
 
565
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i am trying to solve it for b
 

tiny-tim

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Hi Ry122! :smile:

Hint: eliminate θ by using cos²θ + sin²θ = 1. :smile:
 
Move the theta terms in both equations to the left hand side.

Square both equations.

Add the two that you get, now you can use tiny-tim's trick.

What you have left is a quadratic in b, solve it. I'm guessing somewhere along the way you assumed that b > 0 since every else you explicitly took care of the signs, you can use this to throw away a spurious solution if that is the case (leaving you with only one value for b).

Now take either one of the two original equations and plug b into it, and then solve for theta.

Ok that was a complete set of instructions instead of a hint, but this is after all just mathematical gymnastics and not a meaningful exercise.
 
565
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Is this right?
(85.7cos(theta)+85.7sin(theta)=bcos30-85.7+bsin30)^2
7344.49cos^2(theta)+7344.49sin^2(theta)=b^2cos^2(90)-(85.7)^2+b^2sin90
1=(b^2cos^2(90)-(85.7)^2+b^2sin90)/7344.49x7344.49
 
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565
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i cant solve it because theres still a cos^2 and a sin^2. how do i get rid of them?
 

tiny-tim

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Is this right?
(85.7cos(theta)+85.7sin(theta)=bcos30-85.7+bsin30)^2
7344.49cos^2(theta)+7344.49sin^2(theta)=b^2cos^2(90)-(85.7)^2+b^2sin90
1=(b^2cos^2(90)-(85.7)^2+b^2sin90)/7344.49x7344.49
No no no … you added, and then squared. :frown:

(btw, why didn't you just copy my θs and ²s … it makes reading much easier?)

Square first, then add!

Try again! :smile:
 
565
2
(85.7)²cos²θ=b²cos²90-(85.7)²
(85.7)²sin²θ=b²sin²90

(85.7)²cos²θ+(85.7)²sin²θ=b²cos²90-(85.7)²+b²sin²90
this is the same isnt it? or am i adding incorrectly?
 
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tiny-tim

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(85.7)²cos²θ=b²cos²90-(85.7)²
(85.7)²sin²θ=b²sin²90

(85.7)²cos²θ+(85.7)²sin²θ=b²cos²90-(85.7)²+b²sin²90
this is the same isnt it?
No no no … you've put (bcos30 - 85.7)² = (bcos30)² - (85.7)². :cry:

What should it be?

And simplify the left-hand side … that was the whole point … ! :smile:

(oh, and (bcos30)² is b²cos²30. not b²cos²90 :rolleyes:)
 
565
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should it be
bcos30=b.866
(b)²(.866)²

should - (85.7)² be (-85.7)^2?
 
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565
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sorry i understand what you are saying now
 

tiny-tim

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Hi Ry122! :smile:

Are you ok now on this?

If so, hit "[SOLVED]" under "Thread Tools" on the menu bar. :smile:
 
565
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I still need some help.
(85.7cos(theta))²=(bcos30-85.7)² eq1
(85.7sin(theta))²=(bsin30)² eq2
85.7²cos²(theta)=b²cos²30-85.7bcos30-85.7bcos30+85.7² eq1
simplified
85.7²cos²(theta)=b²cos²30-171.4bcos30+85.7²
eq2
(85.7sin(theta))²=(bsin30)²
85.7²sin²(theta)=b²sin²30

add them
85.7²sin²(theta)+85.7²cos²(theta)=b²sin²30+b²cos²30-171.4bcos30+85.7²
sin²(theta)+cos²(theta)=(b²sin²30+b²cos²30-171.4bcos30+85.7²)/(85.7)^4
1=(b²sin²30+b²cos²30-171.4bcos30+85.7²)/(85.7)^4
Is this correct?
I dont know how to solve it with that sin squared and cos squared still in there.
 
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565
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ok i solved it. I got 39.152. Can someone please tell me if this is the correct answer? It would be greatly appreciated.
 
If you have values for both b and theta, then just check it yourself-- plug into both equations, if both equations are valid then you found the correct solution.
 

tiny-tim

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85.7²sin²(theta)+85.7²cos²(theta)=b²sin²30+b²cos²3 0-171.4bcos30+85.7²
sin²(theta)+cos²(theta)=(b²sin²30+b²cos²30-171.4bcos30+85.7²)/(85.7)^4
1=(b²sin²30+b²cos²30-171.4bcos30+85.7²)/(85.7)^4
Is this correct?
Hi Ry122! :smile:

Sorry … no!

First, where did your "^4" come from??

Second, the whole point of adding equal amounts of cos²θ and sin²θ (btw, why did you stop copy-and-pasting the ²θ?) was so that you could replace cos²θ + sin²θ by 1 in the very next line!

Then you get 85.7² on both sides, so they cancel out (in this case … they wouldn't normally).

Try again! :smile:
 
565
2
First, where did your "^4" come from??
i had to get rid of the 2 85.7² on the left side so i moved them to the denominator on the right side


85.7²sin²(theta)+85.7²cos²(theta)=b²sin²30+b²cos²3 0-171.4bcos30+85.7²


85.7² x 85.7² = 85.7^4

sin²(theta)+cos²(theta)=(b²sin²30+b²cos²30-171.4bcos30+85.7²)/(85.7)^4

then use the identity

1=(b²sin²30+b²cos²30-171.4bcos30+85.7²)/(85.7)^4
 
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tiny-tim

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85.7²sin²(theta)+85.7²cos²(theta)=b²sin²30+b²cos²3 0-171.4bcos30+85.7²

85.7² x 85.7² = 85.7^4

sin²(theta)+cos²(theta)=(b²sin²30+b²cos²30-171.4bcos30+85.7²)/(85.7)^4
Nooo … it should only be /(85.7)² at the end! :rolleyes:

Now try it! :smile:
 
565
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can you tell me why?
 

tiny-tim

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erm … 'cos you've got (85.7)² on the left, multiplied by something, plus (85.7)² multiplied by something else … that's only one factor of (85.7)². :smile:

Where did you think the 85.7² x 85.7² came from? :confused:
 
565
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so are you saying this is wrong?
5x²+10x²=15/1
5x²+10=15/1 x x²
5+10=15/1 x x² x x²
 
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tiny-tim

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so are you saying this is wrong?
5x²+10x²=15/1
5x²+10=15/1 x x²
5+10=15/1 x x² x x²
Hi Ry122! :smile:

Yes, and yes². :redface:

(btw, couldn' you have used some letter other than x? :rolleyes:)

Suppose x = 2.

Then 5x²+10x² = 5.4 + 10.4 = 20 + 40 = 60, which we could also write:
5x²+10x² = (5 + 10)x² = 15x² = 15.4 = 60.

If we divide by x², we must divide the whole of each side of the equation by x².

We can't just divide part of it … it simply doesn't work!

(of course, it does work if you put x = 1 ! :smile:)

Then we get (5x²+10x²)/x² = 60/x²,
which obviously we simplify to:
5 + 10 = 60/x², = 60/4 = 15. :smile:

Your way doesn't work: 5x²+10 = 5.4 + 10 = 30, which is not 60/4, and 5 + 10 is not 60/4x4. :redface:
 

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