Finding Solutions for Trigonometric Equations with Small Positive Roots

[skrat]

Homework Statement

I ran across this equation: $$ T_1\cos \vartheta =-T_2\cos \alpha \sin \varphi + T_2\sin \alpha \cos \varphi$$
Get $\alpha$ if $T_1=80$, $T_2=50$, $\vartheta =60^°$ and $\varphi =30^°$.

Homework Equations

The Attempt at a Solution

There are two possible ways:
a) One could simply write $$\sin^2=1-\cos^2\alpha$$ and get a quadratic equation for $\cos \alpha$ with solutions $$(\cos\alpha)_1=0.1196$$ and $$(\cos\alpha)_2=-0.9196.$$ Each solution has two angles $\alpha$. First solution is $$\alpha_1=\pm 83.13^°$$ and second $$\alpha _2=\pm 156.87^°$$

b) One could use some trigonometric identities and rewrite the equation as $$T_1 \cos\vartheta = T_2\sin(\alpha-\varphi)$$ which means (according to my understanding...) $$\alpha =\varphi+arcsin(\frac{T_1}{T_2}\cos\vartheta)+n\pi.$$ For $n=0$ the solution is (as above in case a)) $\alpha_1= 83.13^°$, however the problem here is that I can't figure out how to get the other solution in this case.This is absolutely the most embarrassing question when you realize you are close to a second Bachelor degree...

 

[Samy_A]

$\sin(\pi - x)=\sin(x)$
That should give you the other solution.
Remember that one has to restrict the range of the inverse trigonometric functions in order to have them as well defined functions.

 

[vela]

One could use some trigonometric identities and rewrite the equation as $$T_1 \cos\vartheta = T_2\sin(\alpha-\varphi)$$ which means (according to my understanding...) $$\alpha =\varphi+arcsin(\frac{T_1}{T_2}\cos\vartheta)+n\pi.$$ For $n=0$ the solution is (as above in case a)) $\alpha_1= 83.13^°$, however the problem here is that I can't figure out how to get the other solution in this case.

The $n\pi$ part is wrong. Consider, for example, if $\varphi = 0$ and $\frac{T_1}{T_2}\cos\vartheta = 1$. Then you'd have $\sin \alpha = 1$. One solution is $\alpha = \frac{\pi}{2}$, but clearly $\alpha = -\pi/2$ (n = -1) and $\alpha = 3\pi/2$ (n=1) aren't solutions.

 

[Ray Vickson]

Homework Statement

I ran across this equation: $$ T_1\cos \vartheta =-T_2\cos \alpha \sin \varphi + T_2\sin \alpha \cos \varphi$$
Get $\alpha$ if $T_1=80$, $T_2=50$, $\vartheta =60^°$ and $\varphi =30^°$.

Homework Equations

The Attempt at a Solution

There are two possible ways:
a) One could simply write $$\sin^2=1-\cos^2\alpha$$ and get a quadratic equation for $\cos \alpha$ with solutions $$(\cos\alpha)_1=0.1196$$ and $$(\cos\alpha)_2=-0.9196.$$ Each solution has two angles $\alpha$. First solution is $$\alpha_1=\pm 83.13^°$$ and second $$\alpha _2=\pm 156.87^°$$

b) One could use some trigonometric identities and rewrite the equation as $$T_1 \cos\vartheta = T_2\sin(\alpha-\varphi)$$ which means (according to my understanding...) $$\alpha =\varphi+arcsin(\frac{T_1}{T_2}\cos\vartheta)+n\pi.$$ For $n=0$ the solution is (as above in case a)) $\alpha_1= 83.13^°$, however the problem here is that I can't figure out how to get the other solution in this case.This is absolutely the most embarrassing question when you realize you are close to a second Bachelor degree...

If you have $0 < v < 1$ and an equation $\sin(\theta) = v$, the two smallest positive roots are $\theta_1 = \arcsin(v)$ and $\theta_2 = \pi - \theta_1$.

Just look at the graph of $y = \sin(\theta)$ for $0 \leq \theta \leq \pi$. You will see that the horizontal line $y = v$ cuts the sine-graph at two points, $\theta_1$ and $\pi - \theta_1$. Alternatively, look at the unit circle in the $(x,y)$ plane, and notice that points at angles $\theta$ and $\pi - \theta$ have the same y-coordinate.