351. Prove the equality: $\frac{\sin n\theta}{\cos^n\theta} = {}^nC_1\tan\theta - {}^nC_3\tan^3\theta + {}^nC_5\tan^5\theta - \ldots + A$, where $A = (-1)^{(n - 2)/2}.{}^nC_{n - 1}\tan^{n - 1}\theta$ if $n$ is odd, $A = (-1)^{n/2}.\tan^n\theta$ if $n$ is odd.

Solution: $(\cos\theta + i\sin\theta)^n = \cos^n\theta + i.{}^nC_1\cos^{n - 1}\theta\sin\theta + i^2.{}^nC_2\cos^{(n - 2)}\theta\sin^2\theta + \cdots + i^r.{}^nC_r\cos{(n - r + 1)}\theta\sin^{r - 1}\theta + \cdots$

Separating imaginary patrt part, $\sin n\theta = {}^nC_1\cos^{n-1}\theta\sin\theta - {}^nC_3\cos^{n - 3}\theta\sin^3\theta + \cdots$

Taking into account the parity of $n$ and dividing both members of these equalities by $\cos^n\theta$, we get the required formulas.

352. Prove the following equality: $$ 2^{2m}\cos^{2m}x = \sum_{k = 0}^{m - 1}2\binom{2m}{k}\cos2(m - k)x + \binom{2m}{m}$$

Solution: $\cos\theta = \frac{(\cos\theta + i\sin\theta) + (\cos\theta - i\sin\theta)}{2}$. Let $\cos\theta + i\sin\theta = z$ then $\cos\theta - i\sin\theta = z^{-1} $.

$\therefore \cos^{2m}\theta = \left(\frac{z + z^{-1}}{2}\right)^{2m} = \frac{1}{2^{2m}}\displaystyle\sum_{k=0}^{2m}{}^{2m}C_kz^{2m - k}.z^{-k}$

Moreover $2^{2m}\cos^{2m}\theta = \displaystyle\sum_{k = 0}^{m - 1}{}^{2m}C_kz^{2(m - k)} + {}^{2m}C_m + \sum_{k = m + 1}^{2m}{}^{2m}C_{k}z^{2(m - k)}$

Putting $m - k = -(m - k')$, we rewrite the sum $\displaystyle\sum_{k' = m - 1}^0{}^{2m}C_{2m - k'}z^{-2(m - k')} = \sum_{k = 0}^{m - 1}{}^{2m}C_{k}z^{-2(m - k)}$

And so $\displaystyle 2^{2m}\cos^{2m}\theta = \sum_{k = 0}^{m - 1}{}^{2m}C_k(z^{2(m - k)} + z^{-2(m - k)}) + {}^{2m}C_m$.

However, $z^{2(m - k)} + z^{-2(m - k)} = 2\cos2(m - k)$.

$\therefore\displaystyle 2^{2m}\cos^{2m}\theta = \sum_{k = 0}^{m - 1}2\binom{2m}{k}\cos2(m - k)x + \binom{2m}{m}$.

353. Prove the following equality: $$ 2^{2m}\sin^{2m}x = \sum_{k = 0}^{m - 1}(-1)^{m + k}2\binom{2m}{k}\cos2(m - k)x + \binom{2m}{m}$$

Solution: Putting $\theta = \frac{\pi}{2} - \theta$ in the previous problem, we get the required formula.

354. Prove the following equality: $$ 2^{2m}\cos^{2m + 1}x = \sum_{k = 0}^m2\binom{2m + 1}{k}\cos(2m - 2k + 1)x$$

Solution: This can be deduced like previous problem.

355. Prove the following equality: $$ 2^{2m}\sin^{2m + 1}x = \sum_{k = 0}^m(-1)^{m + k}2\binom{2m+1}{k}\cos(2m - 2k + 1)x$$

Solution: This can be deduced like previous problem.

356. Let $u_n = \cos\alpha + r\cos(\alpha + \theta) + r^2\cos(\alpha + 2\theta) + \ldots + r^n\cos(\alpha + n\theta), v_n = \sin\alpha + r\sin(\alpha + \theta) + r^2\sin(\alpha + 2\theta) + \ldots + r^n\sin(\alpha + n\theta)$, then show that $$ u_n = \frac{\cos\alpha - r\cos(\alpha - \theta) - r^{n + 1}\cos[\alpha + (n + 1)\theta] + r^{n + 2}\cos(\alpha + n\theta)}{1 - 2r\cos\theta + r^2},$$ $$ v_n = \frac{\sin\alpha - r\sin(\alpha - \theta) - r^{n + 1}\sin[\alpha + (n + 1)\theta] + r^{n + 2}\sin(\alpha + n\theta)}{1 - 2r\cos\theta + r^2}$$

Solution: We have the expression $u_n + iv_n = (\cos\alpha + i\sin\alpha) + r[\cos(\alpha + \theta) + i\sin(\alpha + \theta)] + \cdots + r^n[\cos(\alpha + n\theta) + i\sin(\alpha + n\theta)]$

$= (\cos\alpha + i\sin\alpha)[1 + (\cos\theta + i\sin\theta) + \cdots + r^n(\cos n\theta + i\sin n\theta)]$. Putting $z = \cos\theta + i\sin\theta$, then

$u_n + iv_n = (\cos\alpha + i\sin\alpha)[1 + rz + \cdots + r^nz^n] = e^{i\alpha}\frac{(rz)^{n + 1} - 1}{rz - 1}$

Transforming $\frac{(rz)^{n + 1} - 1}{rz - 1}$, separating real part from the imaginary one.

$\frac{(rz)^{n + 1} - 1}{rz - 1} = \frac{[(rz)^{n + 1} - 1][\overline{rz} - 1]}{(rz - 1)(\overline{rz} - 1)}$

$= \frac{r^{n + 2}[\cos n\theta + i\sin n\theta] - r[\cos\theta - i\sin\theta]}{1 - 2r\cos\theta + r^2} + \frac{-r^{n + 1}[\cos(n + 1)\theta + i\sin(n + 1)\theta] + 1}{1 - 2r\cos\theta + r^2}$

Multiplying above with $(\cos\alpha + i\sin\alpha)$ and separating real and imaginary parts we have

$u_n + iv_n = \frac{\cos\alpha - r\cos(\alpha - \theta) - r^{n + 1}\cos[\alpha + (n + 1)\theta] + r^{n + 2}\cos(\alpha + n\theta)}{1 - 2r\cos\theta + r^2} + $

$i\frac{\sin\alpha - r\sin(\alpha - \theta) - r^{n + 1}\sin[\alpha + (n + 1)\theta] + r^{n + 2}\sin(\alpha + n\theta)}{1 - 2r\cos\theta + r^2}$.

Note: Putting $\alpha = 0, r = 1$, we obtain $1 + \cos\theta + \cos2\theta + \cdots + \cos n\theta = \frac{\sin\frac{n + 1}{2}\theta\cos\frac{n\theta}{2}}{\sin\frac{\theta}{2}}$

and $\sin\theta + \sin2\theta + \cdots + \sin n\theta = \frac{\sin\frac{n + 1}{2}\theta\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}$.

357. Simplify the following sums: $$ S = 1 + n\cos\theta + \frac{n(n - 1)}{1.2}\cos2\theta + \ldots = \sum_{k = 0}^n{}^nC_k\cos k\theta,\;[{}^nC_0 = 1]$$ $$ S' = 1 + n\sin\theta + \frac{n(n - 1)}{1.2}\sin2\theta + \ldots = \sum_{k = 0}^n{}^nC_k\sin k\theta,\;[{}^nC_0 = 1]$$

Solution: $S + iS' = \displaystyle\sum_{k=0}^n{}^nC_k(\cos k\theta + i\sin k\theta) = \sum_{k = 0}^n(\cos\theta + i\sin\theta)^k$

$= (1 + \cos\theta + i\sin\theta)^n= \left[2\cos^2\frac{\theta}{2} + 2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}\right]^n = 2^n\cos^n\frac{\theta}{2}\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right)^n$

$= 2^n\cos^n\frac{\theta}{2}\left(\cos\frac{n\theta}{2} + i\sin\frac{n\theta}{2}\right)$. Equating real and imaginary parts we have $S$ and $S'$.

358. If $\alpha = \frac{\pi}{2n}$ and $p < 2n(p\in\mathbb{P})$, then prove that $$ \sin^{2p}\alpha + \sin^{2p}2\alpha + \ldots + \sin^{2p}n\alpha=\frac{1}{2} + n\frac{1.3.5.\ldots(2p - 1)}{2.4.\ldots 2p}$$

Solution: Put $S = \sin^{2p}\alpha + \sin^{2p}2\alpha + \cdots + \sin^{2p}2\alpha = \displaystyle\sum_{l = 1}^n\sin^{2p}l\alpha$

But we have proved earlier $\sin^{2p}l\alpha = \frac{1}{2^{2p - 1}})(-1)^p\displaystyle\sum_{k = 0}^{p - 1}{}^{2p}C_k\cos2(p - k)l\alpha + \frac{1}{2^{2p}}{}^{2p}C_p$, therefore

$S = \frac{(-1)^p}{2^{2p - 1}}\displaystyle\sum_{k = 0}^{p - 1}(-1)^k{}^{2p}C_k\sum_{l = 1}^n\cos2(p - k)l\alpha + \frac{1}{2^{2p}}{}^{2p}C_p$

Put $2(p - k)\alpha = \theta, \sum_{l = 1}^n\cos2(p - k)\alpha = \cos\theta + \cdots + \cos n\theta = \frac{\sin\frac{n\theta}{2}\cos\frac{n + 1}{2}\theta}{\sin\frac{\theta}{2}}$

Denoting $\frac{\sin\frac{n\theta}{2}\cos\frac{n + 1}{2}\theta}{\sin\frac{\theta}{2}} = \sigma_k$, we can prove that $\sigma_k = 0$ if $k$ is of the same parity as $p\{k\equiv p(\mod\;2)\}$ and $\sigma_k = -1$ if $k$ and $p$ are of different parity $\{k\equiv p + 1(\mod)\;2\}$, and we get

$S = \frac{(-1)^{p + 1}}{2^{2p - 1}}\displaystyle\sum_{{k=0, k\equiv p + 1 (\mod 2)}}^{p - 1}(-1)^k{}^{2p}C_k + \frac{n}{2^{2p}}{}^{2p}C_p$.

Hence, $S = \frac{1}{2^{2p - 1}}\displaystyle\sum_{{k=0, k\equiv p + 1 (\mod 2)}}^{p - 1}{}^{2p}C_k + \frac{n}{2^{2p}}{}^{2p}C_p$.

But we can prove that $\displaystyle\sum_{{k=0, k\equiv p + 1 (\mod 2)}}^{p - 1}{}^{2p}C_k = 2^{2p - 2}$ (check binomial theorem chapter) and hence our formula is deduced.

359. Prove that $(x + y)^n - x^n - y^n$ is divisible by $xy(x + y)(x^2 + xy + y^2)$ if $n$ is an odd number and not divisible by $3$.

Solution: Considering the given expression as a polynomial in $y$ we see that at $y = 0$ the polynomial vanishes. Therefore, our polynomial is divisible by $y$. Since it is symmetrical both w.r.t. to $x$ and $y$ this must also be true for $x$ i.e. the polynomial being divisible by $x$. Hence, the polynomial is divisible by $xy$. Putting $y = -x$(we do this for checking divisibility by $x + y$), we have $(x - x)^n - x^n - (-x)^n = 0$. Consequently, the polynomial is divisible by $x + y$.

Now it remains to prove that the polynomial is divisible by $x^2 + xy + y^2$. Expansind this into linear factors we have $x^2 + xy + y^2 = (y - x\omega)(y - x\omega^2)$ where $\omega$ is cube root of unity, which leads to $1 + \omega + \omega^2 = 0$.

Since $n = 3m + 1, 3m + 2\;\forall\;m\in \mathbb{I}$, we substitute $y = x\omega$ and $y = x\omega^2$ and find that it vanishes for both. Consequently, we have proven the divisibility condition.

360. Prove that $(x + y)^n - x^n - y^n$ is divisible by $xy(x + y)(x^2 + xy + y^2)^2$ if $n$, when divided by $6$ has a remainder of $1$.

Solution: Let the quantities $-x, -y$ and $x + y$ be the roots of the cubic equation $x^3 - rx^2 - px - q = 0$. Then $r = -x - y + x + y = 0, -p = xy - x(x + y) - y(x + y), q = xy(x + y)$ reducing our equation to $x^3 - px + q = 0$.

Putting $(x + y)^n - x^n - y^n = S_n$ we find that between successive values of $S_n$ their exists relationship $S_{n + 3} = pS_{n + 1} + qS_n$. We will use mathematical induction to prove that $S_n$ is divisible by $p^2$ with the knowledge that $S_1 = 0$.

Let $S_n$ be divisble by $p^2$ then let $S_{n + 6}$ be also divisible by $p^2$. We have, $S_{n + 6} = pS_{n + 4} + qS_{n + 3}, S_{n + 4} = pS_{n + 2} + qS_{n + 1}$. Therefore,

$S_{n + 6} = p(pS_{n + 2} + qS_{n + 1}) + q(pS_{n + 1} + qS_n) = p^2S_{n + 2} + 2pqS_{n + 1} + q^2S_n$.

Since by supposition, $S_n$ is divisible by $p^2$, it suffices to prove that $S_{n + 1}$ is divisible by $p$. Thus, we only have to prove that given expression is divisible by $x^2 + xy + y^2$ if $n\equiv 2(\mod 6)$, which can be proved by proceeding like previous problem.