91Ó°ÊÓ

Knowing that $$ \cos(n\theta)=\cos^n\theta-\frac{n(n-2)}{2!}\cos^{n-2}\theta+\cdots+(-1)^k\frac{n(n-2)\cdots(n-4k+2)}{(2k)!}\cos^{n-4k}\theta $$ we can express \(\cos(5\theta)\) as $$ \cos(5\theta) = \cos^5\theta - \frac{5\cdot3}{2!}\cos^3\theta + \frac{5\cdot3\cdot1}{4!}\cos\theta $$

We can simplify the above expression by calculating the coefficients of each term: $$ \cos(5\theta) =\cos^5\theta - \frac{15}{2}\cos^3\theta + \frac{15}{24}\cos\theta $$ Denote $$ \cos^5\theta = A, $$ $$ \cos^3\theta = B, $$ and $$ \cos\theta = C $$ we have $$ A = 16C^5, \quad B = 20C^3, \quad C = 5C $$ Multiplying the above expression by \(16\) and substituting \(A\), \(B\), and \(C\), we have $$ 16\cos(5\theta) = 16A - \frac{240}{2}B + \frac{240}{24}C $$

We now write the given equation as $$ 16\cos(5\theta) = 16\cos^5\theta-20\cos^3\theta+5\cos\theta $$ Comparing this with the expression obtained in step 2, we see that $$ 16A - \frac{240}{2}B + \frac{240}{24}C = 16\cos^5\theta-20\cos^3\theta+5\cos\theta $$ Using the equations \(A=16C^5, B=20C^3, C=5C\), we substitute for \(A\), \(B\), and \(C\) and get $$ 16\cdot16C^5 - \frac{240}{2}\cdot20C^3 + \frac{240}{24}\cdot5C = 16\cos^5\theta-20\cos^3\theta+5\cos\theta $$ Simplifying the left-hand side of the equation, we obtain $$ 256C^5 - 2400C^3 + 50C = 16\cos^5\theta-20\cos^3\theta+5\cos\theta $$ Now, divide both sides by \(16\) and substitute back \(C=\cos\theta\): $$ 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta = \cos(5\theta) $$ Therefore, we have shown that $$ \cos(5\theta) = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta $$ for all \(\theta\).