How Can Radian Conversions Simplify Complex Trigonometric Problems?

[Physics lover]

Homework Statement

The question is in Attempt at a solution

Relevant Equations

2CosACosB=Cos(A+B)+Cos(A-B)
CosACos2ACos4A....Cos$2^n$A=Sin($2^n$A)/($2^n$SinA)

I started by converting it into radian.
Then i wrote as $I=Cos(\pi/45)Cos(2\pi/45)...Cos(22\pi/45) Also I=Cos(44\pi/45)Cos(43\pi/45)...Cos(23\pi/45)$
And then i multiplied the two equations.But it didn't helped me.What shall i do further.

 

[BvU]

Please don't write

I=Cos(pi/45)Cos(2pi/45)...Cos(22pi/45)​

but

I=\cos(\pi/45)\cos(2\pi/45)...\cos(22\pi/45):
$$ I=\cos(\pi/45)\cos(2\pi/45)...\cos(22\pi/45): $$​

The question is in Attempt at a solution

Really ! You should know by now that PF doesn't like that !

Homework Equations: $\quad 2\cos A \cos B=\cos(A+B)+\cos(A-B)$
$$\cos A\cos (2A) \cos(4A)...\cos (2^n A)=\sin(2^n A)/(2^n\sin A)$$

I don't see this second equation come back in your attempt... it looks promising, but not that some factors in the product don't appear

​

 

[Physics lover]

Please don't write

I=Cos(pi/45)Cos(2pi/45)...Cos(22pi/45)​

but

I=\cos(\pi/45)\cos(2\pi/45)...\cos(22\pi/45):
$$ I=\cos(\pi/45)\cos(2\pi/45)...\cos(22\pi/45): $$​

Really ! You should know by now that PF doesn't like that !
I don't see this second equation come back in your attempt... it looks promising, but not that some factors in the product don't appear

​

I used in my attempt but it was unsuccessful so ididn't mentioned it.I took out terms from whole series that fits the formula but i cannot proceed further.

 

[Physics lover]

Please somebody answer.I am still trying to do it.

 

[Charles Link]

I'm stuck. You might try a numerical solution to determine what the correct answer may be. I don't presently see a simple way of combining terms. One item of interest is $\cos{60^{\circ}}=1/2$, but attempts to combine the terms to make them look like $\cos{60^{\circ}}$ have eluded me.

 

[Physics lover]

I'm stuck. You might try a numerical solution to determine what the correct answer may be. I don't presently see a simple way of combining terms. One item of interest is $\cos{60^{\circ}}=1/2$, but attempts to combine the terms to make them look like $\cos{60^{\circ}}$ have eluded me.

Thanks for your effort But no worries.I have solved it.i wrote every cos term as sin2A/sinA and the terms cancelled.

 

[Charles Link]

Thanks for your effort But no worries.I have solved it.i wrote every cos term as sin2A/sinA and the terms cancelled.

An interesting problem. And I wrote all the terms out, too, following your post. The cancellation isn't a simple one, in that terms like $\sin(144^{\circ})$ from $\cos(72^{\circ})=\sin(144^{\circ})/(2 \sin(72^{\circ}))$ cancels $\sin(36^{\circ})$ of $\cos(36^{\circ})=\sin(72^{\circ})/(2 \sin(36^{\circ}))$. Surprisingly, every term in the numerator has a match somewhere in the denominator. (Note: $\sin(36^{\circ})=\sin(144^{\circ})$). $\\$ Edit: And for $\cos(60^{\circ})=\sin(120^{\circ})/(2 \sin(60^{\circ}))$, we have $\sin(120^{\circ})=\sin(60^{\circ})$. $\\$ It should be noted that we can't have a simple pairing where e.g. $\cos(88^{\circ}) \cdot \cos(x)=\frac{1}{4}$, because that would mean $\cos(x)$ would need to be greater than 1 in a couple of cases. $\\$ And very good solution @Physics lover $\\$ Edit: Upon further inspection, the doubled angles in the second quadrant, when their mirror image is mapped into the first quadrant, fill in the gaps so that every four degrees there is a term from the numerator, just like in the denominator.