Simultaneous trigonometric equations

In summary, the conversation discusses solving the equation -85.7cos(theta)+bcos30-85.7=0 and -85.7sin(theta)+bsin30=0 for either b or theta. The conversation includes hints on eliminating theta by using cos²θ + sin²θ = 1, and squaring both equations and adding them together to eliminate the theta terms. The final solution for b is 39.152.
  • #1
Ry122
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.
 
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  • #2
Are you trying to solve the equation for b or theta? It's hard to tell.
 
  • #3
i am trying to solve it for b
 
  • #4
Hi Ry122! :smile:

Hint: eliminate θ by using cos²θ + sin²θ = 1. :smile:
 
  • #5
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.
 
  • #6
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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  • #7
i can't solve it because there's still a cos^2 and a sin^2. how do i get rid of them?
 
  • #8
Ry122 said:
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:
 
  • #9
(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 isn't it? or am i adding incorrectly?
 
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  • #10
Ry122 said:
(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 isn't 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:)
 
  • #11
should it be
bcos30=b.866
(b)²(.866)²

should - (85.7)² be (-85.7)^2?
 
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  • #12
sorry i understand what you are saying now
 
  • #13
Hi Ry122! :smile:

Are you ok now on this?

If so, hit "[SOLVED]" under "Thread Tools" on the menu bar. :smile:
 
  • #14
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 don't know how to solve it with that sin squared and cos squared still in there.
 
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  • #15
ok i solved it. I got 39.152. Can someone please tell me if this is the correct answer? It would be greatly appreciated.
 
  • #16
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.
 
  • #17
Ry122 said:
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:
 
  • #18
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+85.7²85.7² 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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  • #19
Ry122 said:
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:
 
  • #20
can you tell me why?
 
  • #21
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:
 
  • #22
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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  • #23
Ry122 said:
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:
 

1. What are simultaneous trigonometric equations?

Simultaneous trigonometric equations are equations that involve multiple trigonometric functions and must be solved simultaneously to find the values of the variables that satisfy both equations.

2. How do you solve simultaneous trigonometric equations?

To solve simultaneous trigonometric equations, you can use algebraic methods such as substitution or elimination, or you can use graphing or trigonometric identities to find the solutions.

3. What are some common strategies for solving simultaneous trigonometric equations?

Some common strategies for solving simultaneous trigonometric equations include converting the equations to a single trigonometric function, using the sum and difference identities, and using the double angle identities.

4. What is the importance of solving simultaneous trigonometric equations?

Solving simultaneous trigonometric equations is important in many applications, such as engineering, physics, and navigation. It allows us to find the values of unknown angles or sides in a triangle and can be used to model and solve real-world problems.

5. What are some common mistakes to avoid when solving simultaneous trigonometric equations?

Some common mistakes to avoid when solving simultaneous trigonometric equations include forgetting to check for extraneous solutions, using the wrong trigonometric identities, and making errors while simplifying the equations.

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