Question

Express as a function of positive acute angles: $\cos \left(240\right)·\sin \left(-110\right)$.

Answer 1

Simplifying the given trigonometric expression.

It is known that, $\cos (180+x)=-\cos x$, as cosine is negative in the third quadrant.

$\sin (180-x)=\sin x$, as sine is positive in the second quadrant.

And $\sin \left(-x\right)=-\sin x$, as sine is negative in the fourth quadrant.

Using the above results in the given expression:

$\begin{array}{rcl}\cos \left(240\right)·\sin (-110)& =& \left[\cos (180+60)\right]·\left[-\sin \left(110\right)\right]\\ & =& \left[-\cos \left(60\right)\right]·\left[-\sin \left(180-70\right)\right]\\ & =& \left[-\cos \left(60\right)\right]·\left[-\sin \left(70\right)\right]\\ & =& \cos \left(60\right)·\sin \left(70\right)\end{array}$

Therefore, the simplified form of the given trigonometric expression as a function of positive acute angles is $\mathbf{cos}\mathbf{}\left(60\right)\mathbf{·}\mathbf{sin}\mathbf{}\left(70\right)$.