Trigonometry reflection problem

[Lindsayyyy]

Hi,

I was about solving a refrection problem and I'm just one step away from the solution. I stuck at some simple problem I guess.

I have:

$$\alpha=60°+\frac1 3 arcsin(\frac1 {1,33} sin(\alpha))$$

and I want to find the alpha but I have troubles to solve this.

I guess it's not that hard, but I'm failing right now

Thanks for your help in advance

 

[I like Serena]

Hi Lindsayyyy!

The problem you have here has afaik no algebraic solution.
If that is what you're looking for, you won't find it.

What you can do, of course, is make an approximation.
That's easy with for instance http://wolframalpha.com.

Cheers!

 

[gsal]

Hhhmm...I gave it a try, and I solved by hand pretty easily.

This was my approach:

Multiply both side of the equation by 3, to eliminate the 3 in the denominator.
Move the 180 from the righ to the left
take the sine of both sides of the equation
you should have

sin(3a - 180) = sin( arcsin( (1/1.33) sin(a) ) )

but the sin(p-180) = -sin(p)

-sin(3a) = sin( arcsin( (1/1.33) sin(a) ) )

and the sin ( arcsin(x) ) = x

-sin(3a) = (1/1.33) sin(a)

and sin(3a) = 3sin(a) - 4(sin(a))^3

-3sin(a) + 4(sin(a))^3 = (1/1.33)sin(a)

put everything on one side of the equation and you have a cubic equation without a quadratic or constant term:

4(sin(a))^3 - (3 + 1/1.33)sin(a) = 0

Now, we simply say

X = sin(a)
A = 4
C = -(3 + 1/1.33)

and we have

AX^3 + CX = 0

or, if we factor one X out:

X ( AX^2 + C ) = 0

So, one solution is X1 = zero, due to the single X...now, we need to solve the roots of the quadratic equation with no B term:

X2 = +sqrt ( -C/A ) = 0.9685
X3 = -sqrt ( -C/A ) = -0.9685

So, from X = sin(a)

a = arcsin(X)
a1 = arcsin(0) = 0 degrees
a2 = arcsin(0.9685) = 75.578 degrees
a3 = arcsin(-0.9685) = -75.578 degrees

and that's it.

 

[willem2]

You still need to check if the solutions work.

You created an extra solution when you replaced

3a - 180 = arcsin( (1/1.33) sin(a) ) with

sin(3a - 180) = sin(arcsin( (1/1.33) sin(a) ))

 

[I like Serena]

Hhhmm...I gave it a try, and I solved by hand pretty easily.

Awww! You're right.
I made a mistake in my calculation and came out with sine's with different arguments that could not be made equal.
Good catch!

You created an extra solution when you replaced ...

Yes. That is right too...

There is an extra condition that -90 < 3a - 180 ≤ 90.
That is, 30 < a ≤ 90.Cheers!

 

[Lindsayyyy]

Thank you very much for help. Nice approach @ gsal, this helped me a lot.