Trigonometric system

**kolg** · February 21st 2010, 05:54 AM

Hi,

I can't solve this trigonometric system in order to y and only appearing Theta(1), h, L and w:

sin(Theta(1))= x/L1

sin(Theta(2))= (w-x)/L2

cos(Theta(1))= y/L1

cos(Theta(2))=(y-h)/L2

L1+L2=L

If necessary you can use sin(theta(1))=sin(theta(2))=w/L

Thanks

**Soroban** · February 21st 2010, 07:20 AM

Hello, kolg!

I can't solve this trigonometric system in order to y
and only appearing Theta(1), h, L and w. . What does that mean?

. . $[1]\;\;\sin(\theta_1) \:=\: \frac{x}{L_1}$

. . $[2]\;\;\sin(\theta_2) \:=\: \frac{w-x}{L_2}$

. . $[3]\;\;\cos(\theta_1) \:=\: \frac{y}{L_1}$

. . $[4]\;\;\cos(\theta_2) \:=\: \frac{y-h}{L_2}$

. . $[5]\;\;L_1+L_2\:=\:L$

If necessary, you can use: $\sin(\theta_1)\,=\,\sin(\theta_2)\,=\,\frac{w}{L}$
This doesn't make sense.

It may be a case of TMI (too much information).

I will assume that we are to solve for $y.$

Square [1]: . $\frac{x^2}{L_1^2} \:=\:\sin^2(\theta_1)$

Square [3]: . $\frac{y^2}{L_1^2} \:=\:\cos^2(\theta_1)$

Add: . $\frac{x^2}{L_1^2} + \frac{y^2}{L_1^2} \;=\;\underbrace{\sin^2(\theta_1) + \cos^2(\theta_1)}_{\text{This is 1}}$

We have: . $\frac{x^2+y^2}{L_1^2} \:=\:1 \quad\Rightarrow\quad w^2 + y^2 \:=\:L_1^2 \quad\Rightarrow\quad y^2 \:=\:L_1^2 - x^2$

Therefore: . $y \;=\;\pm\sqrt{L_1^2 - x^2}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Oh-oh!

I think they want the answer in terms of $L$ . . . and not $L_1$

**kolg** · February 21st 2010, 11:21 AM

Thanks for the reply, but yes I need that in order to L, and that is what I can't figure out

Thanks

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