$ \left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)^{10}= \left(\cos\frac{10 \pi}{4}+i\sin\frac{10 \pi}{4}\right)$ by de Moivre's formula.

ehsan2004 wrote: $ \left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)^{10}= \left(\cos\frac{10 \pi}{4}+i\sin\frac{10 \pi}{4}\right)$ by de Moivre's formula. The question is not to evaluate that expression. It is very easy by De moivre's theorem. But the question is prove the identity by using the evaluation of that expression. I think I got it. Just use De_Moivre's theorem to arrive at $Im\left(\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)^{10}\right)= Im\left(\left(\cos\frac{10 \pi}{4}+i\sin\frac{10 \pi}{4}\right)\right)$ So, $Im\left(\frac{(1+i)^{10}}{32}\right)=\sin\frac{10 \pi}{4}$ $\Longrightarrow \frac{2\dbinom{10}{1}-2\dbinom{10}{3}+\dbinom{10}{5}}{32}=1$ $\Longrightarrow \dbinom{10}{1}-\dbinom{10}{3}+\frac{1}{2}\dbinom{10}{5}=2^{4} $ as required.

We compute the imaginary part of $(\cos\tfrac{\pi}{4}+\sin\tfrac{\pi}{4}i)^{10}$ in two ways. On one hand, $(\cos\tfrac{\pi}{4}+\sin\tfrac{\pi}{4}i)^{10}=\cos\tfrac{10\pi}{4}+\sin\tfrac{10\pi}{4}i=i$ by de Moivre's formula, which implies that $\text{Im}(\cos\tfrac{\pi}{4}+\sin\tfrac{\pi}{4}i)^{10}=1$. On the other hand, $(\cos\tfrac{\pi}{4}+\sin\tfrac{\pi}{4}i)^{10}=(\tfrac{\sqrt{2}}{2}+\tfrac{\sqrt{2}}{2}i)^{10}=\tfrac{1}{2^5}(1+i)^{10}$, which expands by the Binomial Theorem as $$\frac{1}{2^5}\left[\binom{10}{0}+\binom{10}{1}i+\binom{10}{2}i^2+\dots+\binom{10}{10}i^{10}\right].$$Taking the imaginary part, we are left with $$\frac{1}{2^5}\left[\binom{10}{1}i+\binom{10}{3}i^3+\binom{10}{5}i^5+\binom{10}{7}i^7+\binom{10}{9}i^9\right],$$and using $\tbinom{10}{n}=\tbinom{10}{10-n}$ and $i^n=i^{n\pmod{4}}$, the above further simplifies as follows: $$\frac{1}{2^5}\left[2\binom{10}{1}i-2\binom{10}{3}i+\binom{10}{5}i\right]=\frac{1}{2^4}\left[\binom{10}{1}i-\binom{10}{3}i+\frac{1}{2}\binom{10}{5}\right].$$Hence, the desired imaginary part is $\tfrac{1}{2^4}[\tbinom{10}{1}-\tbinom{10}{3}+\tfrac{1}{2}\tbinom{10}{5}]$, and setting this equal to the $1$ from before gives the requested result. $\blacksquare$