Question

Help!!!!!!!!!!!!!!!!​

Answer 1

Answer:

$\cos\alpha\sin\alpha(\cos\alpha-\sin\alpha)$

Step-by-step explanation:

First, simplify each term:

$\sin\left(\dfrac{\pi}{2}+\alpha\right)=\cos \alpha\\ \\\cos \left(\dfrac{\pi}{2}+\alpha\right)=-\sin \alpha\\ \\\cos \left(\alpha-\dfrac{3\pi}{2}\right)=-\sin \alpha\\ \\\sin \left(\dfrac{3\pi}{2}+\alpha\right)=-\cos \alpha$

Then given expression is equivalent to

$\cos ^3\alpha+(-\sin \alpha)^3-(-\sin \alpha)+(-\cos \alpha)\\ \\=\cos ^3\alpha-\sin^3 \alpha+\sin \alpha-\cos \alpha\\ \\=(\cos\alpha-\sin\alpha)(\cos^2\alpha+\cos\alpha\sin\alpha+\sin^2\alpha)-(\cos\alpha-\sin\alpha)\\ \\=(\cos\alpha-\sin\alpha)(1+\cos\alpha\sin\alpha-1)\ \ [\cos^2\alpha+\sin^2\alpha=1]\\ \\=\cos\alpha\sin\alpha(\cos\alpha-\sin\alpha)$