Simultaneous equation involving cos, sin

[atky1224]

Homework Statement

i was solving a engineering four-bar linkage problem until this part and I'm stuck.

I was trying to find out θ3 and θ4 by using the vector loop method and i got the following equations:

50cosθ3 - 45cosθ4 = 39.67
50sinθ3 - 45sinθ4 = 32.58

is there any lead as to how to solve for the values of θ3 and θ4?

Relevant Equations

considered using sin^2 θ + cos^2 θ = 1 but not sure if i am on the right path.

Thanks a lot in advance!

 

[fresh_42]

I assume there is a typo somewhere since the left sides of your equations are the same and the right ones are not. If you are only interested in the solutions then you could use https://www.wolframalpha.com/. If you want to solve it manually, then what are the revised equations?

 

[atky1224]

I assume there is a typo somewhere since the left sides of your equations are the same and the right ones are not. If you are only interested in the solutions then you could use https://www.wolframalpha.com/. If you want to solve it manually, then what are the revised equations?

oh yes, sorry, there is indeed a typo on the second equation, it should be [50sinθ3 - 45sinθ4 = 32.58]. i wanted to solve it manually, but my attempt was stuck at trying to put everything at the right side and leave θ3 on the left side from the first equation, thus making θ3 = arccos[(39.67+45cosθ4)/50]. then i tried substitute θ3 into the second equation and stuck there with my poor mathematic brain...

 

[fresh_42]

\begin{align*}
50\cos \theta_3-45 \cos \theta_4&=39.67 \\
50\sin\theta_3-45\cos\theta_4&=32.58 \quad \text{ times }(-1)\\
\hline \\
50\cos \theta_3-45 \cos \theta_4&=39.67 \\
-50\sin\theta_3+45\cos\theta_4&=-32.58 \quad \text{ add }\\
\hline \\
50\cos \theta_3-50\sin\theta_3&=7.09 \quad \text{ divide by }50\\
\cos \theta_3-\sin\theta_3&=0.1418
\end{align*}
... which yields ...
https://www.wolframalpha.com/input?i=cos+x+-+sin+x+=+0.1418
or
https://www.wolframalpha.com/input?i=cos+x+-+sin+x+=+0.1418+and+0<x<1
... and you can plug in the result to obtain $\theta_4.$

 

[fresh_42]

This is $\theta_3 = 0.6849616481889901703151683129191662012375748774207522750296488810\ldots$ and in degrees of 360° it's $\theta_3= 39,24541157°.$ Thus
$$
\theta_4=\cos^{-1}\left(\dfrac{50\cos 39,24541157° - 39.67}{45}\right) \approx 91,207°
$$

 

[atky1224]

ohh now i see how it is going. please allow me to digest the whole thing. thanks a lot for the guidance

 

[Steve4Physics]

\begin{align*}
50\cos \theta_3-45 \cos \theta_4&=39.67 \\
50\sin\theta_3-45\cos\theta_4&=32.58 \quad \text{ times }(-1)\\
\hline \\
50\cos \theta_3-45 \cos \theta_4&=39.67 \\
-50\sin\theta_3+45\cos\theta_4&=-32.58 \quad \text{ add }\\
\hline \\
50\cos \theta_3-50\sin\theta_3&=7.09 \quad \text{ divide by }50\\
\cos \theta_3-\sin\theta_3&=0.1418
\end{align*}

For information, I think the second equation is incorrect and should be:
$50\sin\theta_3 - 45\sin\theta_4 = 32.58$

 

[fresh_42]

For information, I think the second equation is incorrect and should be:
$50\sin\theta_3 - 45\sin\theta_4 = 32.58$

?

 

[DrClaude]

?

50cosθ3 - 45cosθ4 = 39.67
50sinθ3 - 45sinθ4 = 32.58

 

[Steve4Physics]

?

Because you have written $\cos\theta_4$ instead of $\sin\theta_4$.

 

[fresh_42]

50cosθ3 - 45cosθ4 = 39.67
50sinθ3 - 45sinθ4 = 32.58

Oops!

That ...
\begin{align*}
50\cdot \cos \theta_3 - 45\cdot \cos \theta_4 &= 39.67 \\
50\cdot \sin \theta_3 - 45\cdot \sin\theta_4 &= 32.58
\end{align*}
... makes it more complicated and non-linear as the image on WA ...
https://www.wolframalpha.com/input?i=50*cos(a)+-+45*cos(b)+=+39.67+AND+50*sin(a)+-+45*sin(b)+=+32.58
... shows. In this case there is no other way (that I knew of) than using the numerical algorithmic solution.

The angles on WA are in RAD, so multiply them with $\dfrac{360}{2\pi}$ if you want normal degrees $(\cdot)°$

 

[Steve4Physics]

is there any lead as to how to solve for the values of θ3 and θ4?
Relevant Equations: considered using sin^2 θ + cos^2 θ = 1 but not sure if i am on the right path.

Here’s an outline possible method.

You have 2 equations of the form:
$a\cos A - b\cos B = p$ and
$a\sin A - b\sin B = q$
where $A$ and $B$ are unknown angles and $a, b, p$ and $q$ are known constants.

Eliminate one of the angles, e.g. eliminate $A$ as follows:
$a\cos A = p + b \cos B$
$a\sin A = q + b \sin B$
Square both of these equations and add them. Use '$\sin^2+\cos^2 = 1$'.

With a bit of algebra you get:
$a^2= p^2 + q^2 + b^2 + 2b(p\cos B + q\sin B)$

For neatness let $k = \frac {a^2 -p^2 - q^2 -b^2}{2b}$ giving $p\cos B + q\sin B = k$ which can be solved.

 

[vela]

Homework Statement: i was solving a engineering four-bar linkage problem until this part and I'm stuck.

I was trying to find out θ3 and θ4 by using the vector loop method and i got the following equations:

50cosθ3 - 45cosθ4 = 39.67
50sinθ3 - 45sinθ4 = 32.58

is there any lead as to how to solve for the values of θ3 and θ4?

It looks like these equations correspond to the vector equation $\vec A - \vec B = \vec C$ where $\vec A = (50, \theta_3)$, $\vec B = (45,\theta_4)$, and $\vec C = (C, \theta_C)$, where $C=\sqrt{39.67^2+32.58^2}$ and $\theta_C = \arctan 32.58/39.67$.

You can isolate each vector and then square the resulting equation to get
\begin{align*}
A^2 &= B^2 + C^2 - 2BC\cos(\theta_4-\theta_C) \\
B^2 &= A^2 + C^2 + 2AC\cos(\theta_3-\theta_C) \\
C^2 &= A^2 + B^2 + 2AB\cos(\theta_3-\theta_4)
\end{align*} It's the same thing Steve did above if you note that $p=C \cos\theta_C$ and $q=C\sin\theta_C$ except you avoid some algebra.