Question 1063062
.
<pre>
Exact value of sin(-7pi/12) is


{{{sin(-7pi/12)}}} = {{{sin(-pi/2 - pi/12)}}} = {{{-cos(pi/12)}}}


To find {{{cos(pi/12)}}}, use the formula of half argument for cosines 

    (see the lesson &nbsp;<A HREF= http://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometric-functions-of-half-argument.lesson>Trigonometric functions of half argument</A>&nbsp; in this site):

cos^2(15°) = (1+cos(30°)/2 = {{{(1+sqrt(3)/2)/2}}} = {{{(2+sqrt(3))/4}}}.


Hence,  
cos(15°) = {{{sqrt(2+sqrt(3))/2}}}.


Thus  {{{sin(-7pi/12)}}} = {{{-sqrt(2+sqrt(3))/2}}}.
</pre>

See the lesson 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=>Trigonometric functions of half argument - Examples</A>

in this site.
-------------------------


Regarding the solution by the tutor "math_helper" for this problem, I want to notice that this solution consists and coincides with my:


{{{( - (sqrt(2)/4)(1+sqrt(3))) }}} = {{{-sqrt(2+sqrt(3))/2}}}.


Everybody who squares both sides can check it and convince himself or herself.