Question 1189940
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Problem 1


We can use this identity
{{{cos(A) + cos(B) = 2*cos( (A+B)/2 )*cos( (A-B)/2 )}}}


In this case, A = 38 and B = 106


{{{cos(A) + cos(B) = 2*cos( (A+B)/2 )*cos( (A-B)/2 )}}}


{{{cos(38) + cos(106) = 2*cos( (38+106)/2 )*cos( (38-106)/2 )}}}


{{{cos(38) + cos(106) = 2*cos( 144/2 )*cos( (-68)/2 )}}}


{{{cos(38) + cos(106) = 2*cos( 72 )*cos( -34 )}}}


{{{cos(38) + cos(106) = 2*cos( 72 )*cos( 34 )}}} Recall that cos(-x) = cos(x).



<font color=red>Answer:</font> {{{2*cos( 72 )*cos( 34 )}}}


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Problem 2


We'll use that previous identity, but in reverse.


We have
(A+B)/2 = 234
(A-B)/2 = 125
as the two arguments in the cosines being multiplied.


For each equation, multiply both sides by 2
A+B = 468
A-B = 250


Then add straight down
2A+0B = 718
2A = 718
A = 718/2
A = 359


This leads to:
A+B = 468
359+B = 468
B = 468-359
B = 109



Therefore,
{{{cos(A) + cos(B) = 2*cos( (A+B)/2 )*cos( (A-B)/2 )}}}


{{{cos(359) + cos(109) = 2*cos( (359+109)/2 )*cos( (359-109)/2 )}}}


{{{cos(359) + cos(109) = 2*cos( 234 )*cos( 125 )}}}


{{{2*cos( 234 )*cos( 125 ) = cos(359) + cos(109)}}}


{{{cos( 234 )*cos( 125 ) = (1/2)*(cos(359) + cos(109))}}}


{{{cos( 234 )*cos( 125 ) = (1/2)*cos(359) + (1/2)*cos(109)}}}



<font color=red>Answer:</font> {{{(1/2)*cos(359) + (1/2)*cos(109)}}}
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