Question

Simplify:

\sin(-\alpha -5\pi)\cdotp \cos(\alpha -\dfrac{\pi}{2} )-\tan(\alpha -\dfrac{3\pi}{2})\cdotp \tan(2\pi-\alpha )

Collected in the board: Trigonometry

Steven Zheng Steven Zheng posted 1 month ago


Answer

\sin(-\alpha -5\pi)\cdotp \cos(\alpha -\dfrac{\pi}{2} )-\tan(\alpha -\dfrac{3\pi}{2})\cdotp \tan(2\pi-\alpha )

=-\sin (\alpha +\pi)\cdotp \cos(\dfrac{\pi}{2}-\alpha )+\tan(\dfrac{\pi}{2}-\alpha )\cdotp \tan(-\alpha )

=-\sin \alpha \cdotp \sin \alpha -\cot \alpha \tan \alpha

=-\sin^2\alpha -1


Steven Zheng Steven Zheng posted 1 month ago

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