Answer :

kate200468
[tex]1)\ \ \ sin(- \alpha )=-sin \alpha \\2)\ \ \ cos2 \alpha =1-2sin^2 \alpha \ \ \ \Rightarrow\ \ \ 2sin^2 \alpha =1-cos2 \alpha \\\\ \Rightarrow\ \ \ sin \alpha = \sqrt{\frac{\big{1-cos2 \alpha}}{\big{2}}} \\\\\\sin(-15^0)=-sin15^0=-\sqrt{\frac{\big{1-cos30^0}}{\big{2}}} =-\sqrt{\frac{\big{1- \frac{ \sqrt{3} }{2} }}{\big{2}}} =-\sqrt{\frac{\big{2- \sqrt{3}}}{\big{4}}} =\\\\ =- \frac{1}{2} \sqrt{2- \sqrt{3}} [/tex]

Answer:

[tex]\frac{\sqrt{6} - \sqrt{2} }{4}[/tex] is the answer

Step-by-step explanation:

I just took the test, and did the calculations.