How to calculate θ when you have sin(θ+α)

[Darkmisc]

Homework Statement

The solution to a problem involves calculating theta when you know what sin (theta + alpha) is. Somehow, the solution is able to eliminate alpha so that theta = sin^-1(root 2/root 3) - cos^-1(root 2/root 3)

Relevant Equations

sin(theta + alpha) = root 2/root 3

Hi everyone

I'm having trouble understanding 4c. I'm able to follow most of the solution, but don't know understand the step where

Could someone please explain this step?

The chapter is about trigonometric identities and formulas, but this doesn't fit any of the formulas given. Thanks

 

[Office_Shredder]

This is not a trigonometric identity. In fact, it's not even an identity. There's a lot more going on under the hood here, that I actually don't fully understand.

$\cos(\theta)+\sqrt{2}\sin(\theta)=r\sin(\theta+\alpha)$ has infinitely many solutions for $\alpha$ and $r$. In fact, for any choice of $\alpha$ that doesn't make $\sin(\theta+\alpha)$ zero, you can just divide and solve for a choice of $r$ that works. This solution is claiming that there exists one choice of $r$ and $\alpha$ that make this equation work for every choice of $\theta$. I don't really see why this would be true. It's using that magic value of $\alpha$ to compute $\theta$

Edit to add: I think I get it a bit more, as far as you can say they $1\cos(\theta)+\sqrt{2}\sin(\theta) = (r\sin(\alpha)) \cos(\theta)+ (r\cos(\alpha)) \sin(\theta)$, you just want $r\sin(\alpha)=1$ and $r\cos(\alpha)=\sqrt{2}$. That's two equations in two unknowns, there's at least a decent chance you can get a solution.

 

[anuttarasammyak]

Are you fine starting with
\cos \theta+\sqrt{2} \sin \theta = \sqrt{2}
? Then
\frac{1}{\sqrt{3}}\cos \theta+\frac{\sqrt{2}}{\sqrt{3}} \sin \theta = \frac{\sqrt{2}}{\sqrt{3}}
\sin \alpha \cos \theta + \cos \alpha \sin \theta = \sin (\theta + \alpha) = \frac{\sqrt{2}}{\sqrt{3}}
where
\alpha = \cos^{-1}\frac{\sqrt{2}}{\sqrt{3}}=\sin^{-1}\frac{1}{\sqrt{3}}
which are equivalent.
So
\theta = \sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}-\sin^{-1}\frac{1}{\sqrt{3}}=\frac{\pi}{2}-2\sin^{-1}\frac{1}{\sqrt{3}}
or in rectangle triangle $\triangle$ ABC
\theta = \angle BAC - \angle ACB=\frac{\pi}{2}-2 \angle ACB
where AB=1, BC=$\sqrt{2}$, CA=$\sqrt{3}$ and $\angle$ABC=R.

 

[nuuskur]

Way to hunt a fly with a cannon!

Denote $x:=\theta$. From $\cos ^2x = 2(1-\sin x)^2$ conclude $3\sin ^2x - 4\sin x +1 = 0$. Put $z:=\sin x$, then either $z=1/3$ or $z=1$. Since $0\leqslant x< \pi/2$, one concludes $x=\arcsin (1/3)$.The part in the solution in OP, where one let's $\cos \theta +\sqrt{2}\sin \theta = r\sin (\theta+\alpha)$ is effectively using the intermediate value theorem. The RHS is parametrised such that it contains $\theta$. And then further effort is applied to determine the parametrisation. It is sufficient, but unnecessary. On the other hand, it's good for practicing technique.

 

[anuttarasammyak]

Thanks for brilliant answer. Just for checking the coincidence with #3,
\cos \theta = \sin(2 \sin^{-1}\frac{1}{\sqrt{3}})=2\frac{1}{\sqrt{3}}\frac{\sqrt{2}}{\sqrt{3}}=\frac{2\sqrt{2}}{3}
\sin\theta = \frac{1}{3}

 

[SammyS]

Another approach at finding a solution for $\displaystyle \ \frac{\cos\theta}{1-\sin\theta}=\sqrt{2} \ :$

Square the equation. Then change $\ \cos^2 \theta \$ to $\ 1-\sin^2 \theta \$ which is equal to $\ (1+\sin \theta )(1-\sin \theta)$ .

After cancelling a factor of $\ (1-\sin \theta) \$ from numerator & denominator we get :

$\displaystyle \frac{1+\sin \theta }{1-\sin \theta}= 2$ .

Solving for $\sin \theta$ we get $\ \displaystyle \sin \theta = \frac 1 3 \$ .