1. ## [SOLVED] Applying Identities

Hello,

Was hoping I could get some direction in this one, can't seem to get the desired results.

The problem:

$2 \cos( \alpha /2 ) = \cos \alpha + 1$ Solve for all possible

Now, using the half angle identity I get

$2 (\pm \sqrt \frac{1+\cos \alpha}{2}) = \cos \alpha + 1$

$4 \frac{1 + \cos \alpha}{2} = (\cos \alpha + 1)^2$

$2 + 2 \cos \alpha = \cos^2 \alpha + 2 \cos \alpha + 1$ simplify

$\cos^2 \alpha - 1 = 0$

This is where I'm at. The only thing I can think of is to factor out a $\cos \alpha$

but then I'm left with $\cos \alpha ( \cos \alpha + \frac{1}{\cos \alpha}) = 0$

or $\cos \alpha ( \cos \alpha + \sec \alpha)= 0$

The answers to the problem are:

$180 + k (720), 540 + k(720), k(720)$

So I'm pretty sure I'm missing something here.

thank you!

2. From $\cos^2 \alpha - 1 = 0$ you get $\cos^2 \alpha = 1$, from which you have the solutions

$\cos \alpha = \pm 1$.

Thus, $\alpha = k \pi$.

The only way you could get three solutions would be to have an equation with $\cos^3 (\alpha + \delta)$ or $\cos (3\alpha + \delta)$ in it, for some $\delta$. Also, the solutions would be equally spaced, which your given solution is not.

3. Put $\alpha=2t$ and your equation becomes $2\cos t=\cos(2t)+1=2\cos^2t,$ which is not hard to solve.

4. (Post removed because I can't read... )

01

5. Originally Posted by icemanfan
From $\cos^2 \alpha - 1 = 0$ you get $\cos^2 \alpha = 1$, from which you have the solutions

$\cos \alpha = \pm 1$.

Thus, $\alpha = k \pi$.

The only way you could get three solutions would be to have an equation with $\cos^3 (\alpha + \delta)$ or $\cos (3\alpha + \delta)$ in it, for some $\delta$. Also, the solutions would be equally spaced, which your given solution is not.
The answer in book says it's it's k(720) or k( $4 \pi$). Instead of $k \pi
$

Because of the original $2 \cos( \alpha /2 )$ the period is 4 $\pi$.
Would that explain why there are three separate answers?

6. Now that I looked at the problem more closely, the problem with your method, which icemanfan and I expanded on...
Originally Posted by icemanfan
From $\cos^2 \alpha - 1 = 0$ you get $\cos^2 \alpha = 1$, from which you have the solutions

$\cos \alpha = \pm 1$.

Thus, $\alpha = k \pi$.
... is that by squaring both sides you introduced an extraneous roots. Saying that the solution is $\alpha = k \pi$ implies that $\alpha = 2\pi$ would be a solution, but plugging in that solution for alpha doesn't work:
$2 \cos( \alpha /2 ) = \cos \alpha + 1$
$2 \cos( 2\pi /2 ) = \cos 2\pi + 1$
$2(-1) = 1 + 1$
$-2 = 2 ????$

(This is what I get for reading too fast. )

So, Krizalid's method is probably the best one:
Let $\alpha = 2t$, then
\begin{aligned}
2\cos t &= \cos (2t) + 1 \\
2\cos t &= 2\cos^2(2t) - 1 + 1 \\
2\cos t &= 2\cos^2 t \\
2\cos^2 t - 2\cos t &= 0 \\
2\cos t(cos t - 1) &= 0 \\
\end{aligned}

So
2 cos t = 0 or cos t = 0, which means that
t = π/2 + 2πk and t = 3π/2 + 2πk
But we said α = 2t, so
α = π + 4πk and α = 3π + 4πk
[or α = 180° + (720°)k or α = 540° + (720°)k].

cos t - 1 = 0 or cos = 1, which means that
t = 0 + 2πk or t = 2πk
But again, α = 2t, so
α = 4πk
[or α = (720°)k].

01

7. Originally Posted by yeongil
Now that I looked at the problem more closely, the problem with your method, which icemanfan and I expanded on...

... is that by squaring both sides you introduced an extraneous roots. Saying that the solution is $\alpha = k \pi$ implies that $\alpha = 2\pi$ would be a solution, but plugging in that solution for alpha doesn't work:
$2 \cos( \alpha /2 ) = \cos \alpha + 1$
$2 \cos( 2\pi /2 ) = \cos 2\pi + 1$
$2(-1) = 1 + 1$
$-2 = 2 ????$

(This is what I get for reading too fast. )

So, Krizalid's method is probably the best one:
Let $\alpha = 2t$, then
\begin{aligned}
2\cos t &= \cos (2t) + 1 \\
2\cos t &= 2\cos^2(2t) - 1 + 1 \\
2\cos t &= 2\cos^2 t \\
2\cos^2 t - 2\cos t &= 0 \\
2\cos t(cos t - 1) &= 0 \\
\end{aligned}

So
2 cos t = 0 or cos t = 0, which means that
t = π/2 + 2πk and t = 3π/2 + 2πk
But we said α = 2t, so
α = π + 4πk and α = 3π + 4πk
[or α = 180° + (720°)k or α = 540° + (720°)k].

cos t - 1 = 0 or cos = 1, which means that
t = 0 + 2πk or t = 2πk
But again, α = 2t, so
α = 4πk
[or α = (720°)k].

01
Fantastic! Thank you everyone for the help