Question 1181851
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I'll use A and B in place of 𝛼 and 𝛽


If sin(A) = 5/6, then cos(A) = -sqrt(11)/6. You can use the pythagorean trig identity cos^2+sin^2 = 1 to see why this works. Or you can draw out a right triangle to find the missing side (using the pythagorean theorem). So it's not a coincidence that the term 'pythagorean' comes up with this trig identity.


Note: cosine is negative in Q2


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If tan(B) = 1/2, then we have an opposite side of 1 and adjacent side of 2. 
This leads to a hypotenuse sqrt(1^2+2^2) = sqrt(5), which is again where the pythagorean theorem comes up.


Through a bit of algebra, you should find these two facts:
sin(B) = -sqrt(5)/5
cos(B) = -2sqrt(5)/5


Both sine and cosine are negative in Q3


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In the previous two sections, we found the following four facts
cos(A) = -sqrt(11)/6
cos(B) = -2sqrt(5)/5
sin(A) = 5/6
sin(B) = -sqrt(5)/5


Use those four items in the identity below
{{{cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B)}}}


{{{cos(A+B) = (-sqrt(11)/6)*(-2*sqrt(5)/5) - (5/6)*(-sqrt(5)/5)}}}


{{{cos(A+B) = 2*sqrt(11*5)/(6*5) + 5*sqrt(5)/(6*5)}}}


{{{cos(A+B) = 2*sqrt(55)/30 + 5*sqrt(5)/30}}}


{{{cos(A+B) = (2*sqrt(55) + 5*sqrt(5))/30}}}
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