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The problem is as follows:

$\textrm{Find B:}$

$$B=\cos^{3}\omega\sin\omega-\sin^{3}\omega\cos\omega+\frac{\sqrt{3}}{\sin20^{\circ}}-\frac{\sqrt{3}}{\cos20^{\circ}}-4$$

$\textrm{when}\;\omega=4^{\circ}$

The third and fourth terms in the expression can be arranged in a different way but since both are $\sqrt{3}$ there is no way that it can be "transformed" into a sum of angles for sines and cosines. Needless to say that $4^{\circ}$ is not an important angle. What would be the best way to solve this problem?.

Chris Steinbeck Bell
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1 Answers1

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Hint

At least, use

$$\cos^{3}(\omega)\sin(\omega)-\sin^{3}(\omega)\cos(\omega)=\frac 14 \sin(4\omega)$$ $$\frac 1 {\sin(x)}-\frac 1 {\cos(x)}=\frac {\cos(x)-\sin(x)}{\sin(x)\cos(x)}=2\sqrt 2\,\frac {\cos(x+\frac \pi 4)}{\sin(2x)}$$

Claude Leibovici
  • 260,315
  • Although it is some hidden I've reached to the answers you proposed by using power reduction formulas and sum of cosines (not sure if you used them though). However, the final answer is not within the alternatives in my book. Which are $\frac{\sin16^{\circ}}{4}; \frac{\sin16^{\circ}}{2}; \frac{\sin32^{\circ}}{4};2sin16^{\circ};or;\frac{sin32^{\circ}}{2}$ Could it be one of those since it appears $4$ in the denominator?. – Chris Steinbeck Bell Oct 20 '17 at 03:39
  • If the choices are $\frac{\sin16^{\circ}}{4}$, $ \frac{\sin16^{\circ}}{2}$,$\frac{\sin32^{\circ}}{4}$, $2\sin16^{\circ}$, $\frac{\sin32^{\circ}}{2}$, this is not possible since the value of your original expression is negative ! – Claude Leibovici Oct 20 '17 at 03:53
  • How do I know the value is negative?. Other than just pluggin the values in the expression and using a software or a calculator?. – Chris Steinbeck Bell Oct 20 '17 at 08:22