What is the Inverse Cosine of a Squared Angle?

[Mechaman]

Not sure if I'm doing this right. I have an angle theta to find but the cosine has been squared. I brought over inverse cosine to multiply to leave theta on its own. I was told the answer should be closer to 37 degrees? Am I doing something wrong here?

 

[jedishrfu]

Did you calculate it via

arccos(sqrt(50/79.57) ?

 

[Mark44]

Not sure if I'm doing this right. I have an angle theta to find but the cosine has been squared. I brought over inverse cosine to multiply to leave theta on its own. I was told the answer should be closer to 37 degrees? Am I doing something wrong here?

View attachment 89616

On the 4th line from the bottom you have ##\frac{\sigma_n}{\sigma_x} = \cos^2(\theta)##
This is equivalent to ##\cos(\theta) = \pm \sqrt{\frac{\sigma_n}{\sigma_x}}##
If you take the inverse cosine of both sides, you can isolate ##\theta##. You are NOT multiplying by inverse cosine to get this.

There are many values of ##\theta## that satisfy the last equation above. One that I get is around 37.6°. Please show us the calculation you did.

 

[jedishrfu]

Remember the notation ##cos^{-1}(\theta)## is not the same as ##1/cos(\theta)##.

 

[Mechaman]

On the 4th line from the bottom you have ##\frac{\sigma_n}{\sigma_x} = \cos^2(\theta)##
This is equivalent to ##\cos(\theta) = \pm \sqrt{\frac{\sigma_n}{\sigma_x}}##
If you take the inverse cosine of both sides, you can isolate ##\theta##. You are NOT multiplying by inverse cosine to get this.

There are many values of ##\theta## that satisfy the last equation above. One that I get is around 37.6°. Please show us the calculation you did.

Thanks for helping I worked it out as the same 37.56 degrees, my maths is a little rusty at the moment!

 

[Mark44]

Thanks for helping I worked it out as the same 37.56 degrees, my maths is a little rusty at the moment!

Keep in mind that +/-. There's another value around 142°.