Question 1140837
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<pre>
(1)  108° = 90° + 18°.


     Use the trigonometric identity


         sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b).


     You will get


         sin(108°) = sin(90°+18°) = sin(90°)*cos(18°) + cos(90°)*sin(18°).   (*)


    Substitute here  sin(90°) = 1;  cos(90°) = 0;  sin(18°) = k  and  cos(18°) = {{{sqrt(1-sin(18^o))}}} = {{{sqrt(1-k^2)}}}.


    You will get then (by continuing (*) )


         sin(108°) = {{{1*sqrt(1-k^2)}}} + {{{0*k}}} = {{{sqrt(1-k^2)}}}.     <U>ANSWER</U>



2.  cos(-36°)


    cos(-36°) = cos(36°) = cos(2*18°).


    Use the trigonometric identity


        cos(2a) = 1 - 2*sin^2(a).


    You will get then


        cos(-36°) = cos(36°) = cos(2*18°) = {{{1 - 2*k^2}}}.      <U>ANSWER</U>
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For the list of basic trigonometric identities, see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Compendium-of-Trigonometry-Formulas.lesson>FORMULAS FOR TRIGONOMETRIC FUNCTIONS</A>

in this site.