Đáp án:
$P=-\sqrt{3}\sin x\cos x-\dfrac{1}{2}$
Giải thích các bước giải:
$P=\cos(30^{\circ}-x)\cos(90^{\circ}+x)-\sin(30^{\circ}+x)\sin(90^{\circ}-x)\\
=\cos(30^{\circ}-x).\left [ -\sin x \right ]-\sin(30^{\circ}+x)\cos x\\
=-(\cos30^{\circ}\cos x+\sin30^{\circ}\sin x).\sin x-\\ \cos x \left ( \sin30^{\circ}\cos x+\cos30^{\circ}\sin x \right )\\
=-(\dfrac{\sqrt{3}}{2}\cos x+\dfrac{1}{2}\sin x).\sin x-\cos x\left ( \dfrac{1}{2}\cos x+\dfrac{\sqrt{3}}{2}\sin x \right )\\
=-\dfrac{\sqrt{3}}{2}\sin x\cos x-\dfrac{1}{2}\sin^2x-\dfrac{1}{2}\cos^2x-\dfrac{\sqrt{3}}{2}\sin x\cos x\\
=-\sqrt{3}\sin x\cos x-\dfrac{1}{2}(\sin^2x+\cos^2x)\\
=-\sqrt{3}\sin x\cos x-\dfrac{1}{2}$
