If sin a+sin b=-14/65 and cos a+cos b=-8/65, then what is tan(a+b)?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Since the trigonometric functions from the given sums are matching, we'll transform the given sums into products.

sin a + sin b = 2sin[(a+b)/2]*cos[(a-b)/2] (1)

cos a + cos b = 2cos[(a+b)/2]*cos[(a-b)/2] (2)

We'll divide (1) by (2):

(sin a + sin b)/(cos a + cos b)=2sin[(a+b)/2]*cos[(a-b)/2]/2cos[(a+b)/2]*cos[(a-b)/2]

We'll simplify and we'll get:

(sin a + sin b)/(cos a + cos b)=sin[(a+b)/2]/cos[(a+b)/2]

(sin a + sin b)/(cos a + cos b)=tan[(a+b)/2]

But, (sin a + sin b)/(cos a + cos b) = (-14/65)/(-8/65)

(sin a + sin b)/(cos a + cos b) = 14/8

(sin a + sin b)/(cos a + cos b) = 7/4 =>

=> tan[(a+b)/2] = 7/4

We'll determine tan(a+b) = tan 2[(a+b)/2] = 2tan[(a+b)]/2/1-{tan[(a+b)/2]}^2

tan(a+b) = 2*(7/4)/(1 - 49/16)

tan(a+b) = (7/2)/(-33/16)

tan(a+b) = -56/33

The value of tan(a+b) is tan(a+b) = -56/33.

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