Question

use properties of the trigonometric functions to find the exact value of each expression. Do not use a calculator:

$\sin \left(-\frac{\pi }{12}\right)\csc \left(\frac{25\pi }{12}\right)$

Short answer

The exact value of$\sin \left(-\frac{\pi }{12}\right)\csc \left(\frac{25\pi }{12}\right)is-1$

Step-by-step solution

Given information

We are given a trigonometric equation$\sin \left(-\frac{\pi }{12}\right)\csc \left(\frac{25\pi }{12}\right)$

use the periodicity of trigonometric function cscθ+2π=cscθ

We get

$$\begin{gathered} \sin \left(-\frac{\pi }{12}\right)\csc \left(\frac{25\pi }{12}\right) \\ -\sin \left(\frac{\pi }{12}\right)\csc \left(2\pi +\frac{\pi }{12}\right) \\ -\sin \left(\frac{\pi }{12}\right)\csc \left(\frac{\pi }{12}\right) \\ -\sin \left(\frac{\pi }{12}\right)\frac{1}{\sin \left({\displaystyle \frac{\pi }{12}}\right)} \\ =-1 \end{gathered}$$

Conclusion

The exact value of$\sin \left(-\frac{\pi }{12}\right)\csc \left(\frac{25\pi }{12}\right)is-1$