If $\sum_{n=0}^{\infty} \cos ^{2 n} heta=5$, what is $\cos 2 heta$ ?

Question

If $\sum_{n=0}^{\infty} \cos ^{2 n} 	heta=5$, what is $\cos 2 	heta$ ?

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We start with the given series: $\sum_{n=0}^{\infty} \cos^{2n} heta = 5$. ‖ ‖ step 2 ⋮ Recognize that this is a geometric series with the first term $a = 1$ and common ratio $r = \cos^2 heta$. ‖ ‖ step 3 ⋮ The sum of an infinite geometric series is given by $\frac{a}{1 - r}$, so we have $\frac{1}{1 - \cos^2 heta} = 5$. ‖ ‖ step 4 ⋮ Simplify the equation: $1 - \cos^2 heta = \frac{1}{5}$. ‖ ‖ step 5 ⋮ Recognize that $1 - \cos^2 heta = \sin^2 heta$, so $\sin^2 heta = \frac{1}{5}$. ‖ ‖ step 6 ⋮ Therefore, $\cos^2 heta = 1 - \frac{1}{5} = \frac{4}{5}$. ‖ ‖ step 7 ⋮ We need to find $\cos 2 heta$. Recall the double-angle formula: $\cos 2 heta = 2 \cos^2 heta - 1$. ‖ ‖ step 8 ⋮ Substitute $\cos^2 heta = \frac{4}{5}$ into the formula: $\cos 2 heta = 2 \left(\frac{4}{5} ight) - 1 = \frac{8}{5} - 1 = \frac{3}{5}$. ‖ ∻Answer∻ ⚹ $\cos 2 heta = \frac{3}{5}$ ⚹ ∻Key Concept∻ ⚹Geometric Series Sum Formula⚹ ∻Explanation∻ ⚹The sum of an infinite geometric series with first term $a$ and common ratio $r$ (where $|r| < 1$) is given by $\frac{a}{1 - r}$. This formula was used to solve for $\cos^2 heta$ and subsequently $\cos 2 heta$.⚹◊What is the formula for $\cos 2 heta$ in terms of $\cos^{2n} heta$?⍭ Generate me a similar question◊

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