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Post by Micah Pietraho on Jun 6, 2018 18:59:16 GMT
Given that $2^{-\frac{3}{4} + \cos \theta} + 1 = 2^{-\frac{1}{2} + 2\cos \theta}$, find $\cos 2\theta$.
I solved this in about a minute just from an educated guess and knowing the cosine double angle formula. With only a quick attempt at finding a slightly more legitimate solution, I wasn't able to; does anyone have a method other than just noticing the obvious exponents and solving for cos(theta)?
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2017 #5
Jun 6, 2018 21:28:10 GMT
via mobile
Post by Guest on Jun 6, 2018 21:28:10 GMT
Try a variable substitution and transform the equation into a quadratic. Let y = 2^cos(@), jam it through the quadratic formula?
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Post by Admin on Jun 7, 2018 17:08:34 GMT
There are several options, I let x = 1/4 + cos(theta) and it turns into a nice quadratic equation...
you still need a double angle formula, but it gives a solution to cos(theta) without just using inspection.
Mike F.
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Noble Mushtak
New Member
11th Grade for 2017-18; Placed 5th in MAML Regular Season
Posts: 28
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Post by Noble Mushtak on Jun 19, 2018 13:42:06 GMT
I think the first exponent should actually be $-\frac{3}{2}+2\cos\theta$ and the exponent on the right-hand side should be $1/4+\cos\theta$. This is a more complicated method, but I tried squaring both sides:
$$2^{-3+4\cos\theta}+1+2\cdot 2^{-3/2+2\cos\theta}=2^{1/2+2\cos\theta}$$
Simplify the third term on the left-hand side:
$$2^{-3+4\cos\theta}+1+2^{-1/2+2\cos\theta}=2^{1/2+2\cos\theta}$$
Now, the third term on the left-hand side is clearly half of the right-hand side since their exponents differ by 1, so if we subtract the third term on the left-hand side, we get:
$$2^{-3+4\cos\theta}+1=\frac{2^{1/2+2\cos\theta}}{2}$$
In short, if the first equation was $x+1=y$, this equation is $x^2+1=y^2/2$. This gives us:
$$x^2+1=(x+1)^2/2 \rightarrow 2x^2+2=x^2+2x+1 \rightarrow x^2-2x+1=0 \rightarrow (x-1)^2=0 \rightarrow x=1$$
Thus, $x=2^{-3/2+2\cos\theta}=1$. Take the log of both sides to solve for $\cos\theta$ and then apply double-angle formula to find $\cos(2\theta)$.
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