Simultaneous trigonometric equations

1. Apr 10, 2008

Ry122

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.

2. Apr 11, 2008

Snazzy

Are you trying to solve the equation for b or theta? It's hard to tell.

3. Apr 11, 2008

Ry122

i am trying to solve it for b

4. Apr 11, 2008

tiny-tim

Hi Ry122!

Hint: eliminate θ by using cos²θ + sin²θ = 1.

5. Apr 11, 2008

DavidWhitbeck

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. Apr 11, 2008

Ry122

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

Last edited: Apr 11, 2008
7. Apr 11, 2008

Ry122

i cant solve it because theres still a cos^2 and a sin^2. how do i get rid of them?

8. Apr 11, 2008

tiny-tim

No no no … you added, and then squared.

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

Try again!

9. Apr 11, 2008

Ry122

(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?

Last edited: Apr 11, 2008
10. Apr 11, 2008

tiny-tim

No no no … you've put (bcos30 - 85.7)² = (bcos30)² - (85.7)².

What should it be?

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

(oh, and (bcos30)² is b²cos²30. not b²cos²90 )

11. Apr 11, 2008

Ry122

should it be
bcos30=b.866
(b)²(.866)²

should - (85.7)² be (-85.7)^2?

Last edited: Apr 11, 2008
12. Apr 11, 2008

Ry122

sorry i understand what you are saying now

13. Apr 12, 2008

tiny-tim

Hi Ry122!

Are you ok now on this?

14. Apr 12, 2008

Ry122

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

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.

Last edited: Apr 12, 2008
15. Apr 12, 2008

Ry122

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. Apr 12, 2008

DavidWhitbeck

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. Apr 12, 2008

tiny-tim

Hi Ry122!

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!

18. Apr 12, 2008

Ry122

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

Last edited: Apr 12, 2008
19. Apr 12, 2008

tiny-tim

Nooo … it should only be /(85.7)² at the end!

Now try it!

20. Apr 12, 2008

Ry122

can you tell me why?