Solution:
$\frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+2\cos4\alpha+\cos3\alpha}=\frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+\cos3\alpha+2\cos4\alpha}$
$=\frac{2\sin\alpha\cos4\alpha}{2\cos4\alpha\cos\alpha+2\cos4\alpha}$
$=\frac{2\sin\alpha\cos4\alpha}{2\cos4\alpha(\cos\alpha+1)}$
$=\frac{\sin\alpha}{\cos\alpha+1}$
$=\frac{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}+\sin^2\frac{\alpha}{2}+\cos^2\frac{\alpha}{2}}$
$=\frac{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2\cos^2\frac{\alpha}{2}}$
$=\frac{\sin\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}}$
$=\tan\frac{\alpha}{2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
The real value of $\alpha$ for which the expression $\frac{1-\text{i}\sin\alpha}{1+2\text{i}\sin\alpha}$ is purely real is:
$(\text{n}+1)\frac{\pi}{2}$
$(2\text{n}+1)\frac{\pi}{2}$
$\text{n}\pi$
None of these, where $\text{n}\in\text{N}$
Which of the following is not a negation of “A natural number is greater than zero”.