Question 1159222
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I'm not sure what part b is trying to say because the notation is a bit strange. 


I'll show you how to do part a


{{{cos(x/2) = sqrt( (1+cos(x))/2)}}} Half angle identity for cosine, when cosine is positive. Keep in mind that cos(15) is a positive value as the angle 15 degrees is in Q1 where cosine is positive.


{{{cos(30/2) = sqrt( (1+cos(30))/2)}}} Plug in x = 30


{{{cos(15) = sqrt( (1+sqrt(3)/2)/2)}}} Replace cos(30) with sqrt(3)/2


{{{cos(15) = sqrt( (2/2+sqrt(3)/2)/2)}}} Rewrite 1 as 2/2


{{{cos(15) = sqrt( ((2+sqrt(3))/2)/2)}}} Combine the upper fractions


{{{cos(15) = sqrt( ((2+sqrt(3))/2)*(1/2))}}} 


{{{cos(15) = sqrt( (2+sqrt(3))/4 )}}}


{{{cos(15) = sqrt( 2+sqrt(3) )/sqrt(4)}}} Break up the square root


{{{cos(15) = sqrt( 2+sqrt(3) )/2}}} Simplify sqrt(4) into 2. This is as simplified as it gets.



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Side note: if you use a calculator, you should find
{{{cos(15) = 0.96592582628907}}} (make sure your calculator is in degree mode)


{{{sqrt( 2+sqrt(3) )/2 = 0.96592582628907}}}


both are approximations. Since we get the same approximation, this helps confirm the answer.


Another way to confirm the answer is to compute {{{( sqrt(2+sqrt(3)) )/(2)-cos(15)}}} and you should get zero as a result (or some very small number due to rounding error). I'm using the idea that if {{{x = y}}}, then {{{x-y = 0}}}
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