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

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In summary, the conversation discusses a step in solving a problem involving trigonometric identities and formulas. The equation given, ##\cos(\theta)+\sqrt{2}\sin(\theta)=r\sin(\theta+\alpha)## has infinitely many solutions for ##\alpha## and ##r##. The solution in question claims to find one choice of ##r## and ##\alpha## that works for every value of ##\theta##, using the intermediate value theorem. Another approach is discussed, using the equation ##\displaystyle \ \frac{\cos\theta}{1-\sin\theta}=\sqrt{2} \ ## and solving for ##\sin\theta##, with the solution being ##\sin
  • #1
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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

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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

image_2022-04-25_154208747.png
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image_2022-04-25_153916012.png
 
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  • #2
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.
 
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  • #3
Are you fine starting with
[tex]\cos \theta+\sqrt{2} \sin \theta = \sqrt{2}[/tex]
? Then
[tex]\frac{1}{\sqrt{3}}\cos \theta+\frac{\sqrt{2}}{\sqrt{3}} \sin \theta = \frac{\sqrt{2}}{\sqrt{3}}[/tex]
[tex]\sin \alpha \cos \theta + \cos \alpha \sin \theta = \sin (\theta + \alpha) = \frac{\sqrt{2}}{\sqrt{3}}[/tex]
where
[tex]\alpha = \cos^{-1}\frac{\sqrt{2}}{\sqrt{3}}=\sin^{-1}\frac{1}{\sqrt{3}}[/tex]
which are equivalent.
So
[tex]\theta = \sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}-\sin^{-1}\frac{1}{\sqrt{3}}=\frac{\pi}{2}-2\sin^{-1}\frac{1}{\sqrt{3}}[/tex]
or in rectangle triangle ##\triangle## ABC
[tex]\theta = \angle BAC - \angle ACB=\frac{\pi}{2}-2 \angle ACB [/tex]
where AB=1, BC=##\sqrt{2}##, CA=##\sqrt{3}## and ##\angle##ABC=R.
 
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  • #4
Way to hunt a fly with a cannon! :oops:

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.
 
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  • #5
Thanks for brilliant answer. Just for checking the coincidence with #3,
[tex]\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}[/tex]
[tex]\sin\theta = \frac{1}{3}[/tex]
 
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  • #6
image_2022-04-25_154208747-png.png

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 \ ## .
 
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FAQ: How to calculate θ when you have sin(θ+α)

How do I calculate θ when given sin(θ+α)?

To calculate θ, you can use the inverse sine function (sin^-1) on both sides of the equation. This will isolate θ and give you the value in radians. Remember to convert to degrees if necessary.

Can I use a calculator to find the value of θ?

Yes, most scientific calculators have a function for inverse sine (sin^-1) which can be used to find the value of θ when given sin(θ+α).

Is there a specific formula for calculating θ from sin(θ+α)?

Yes, the formula is θ = sin^-1(sin(θ+α)). However, this may vary depending on the context and problem you are working on.

What if I am given a decimal value for sin(θ+α)?

If you are given a decimal value for sin(θ+α), you can use a scientific calculator to find the inverse sine (sin^-1) of that value and solve for θ.

Are there any special cases or restrictions when solving for θ from sin(θ+α)?

Yes, when using the inverse sine function, the value inside the parentheses (θ+α) must be between -1 and 1. If it is not, there is no solution for θ. Additionally, θ may have multiple solutions depending on the given value of α.

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