Sine and Tangent Identities

Question

If cos A = 1/2,  sin B =  (-3/5) where A and B arein Q IV, find:

~               cos (A + B)          (I calculated with the formula 4 -3√3 / 10 )

~                The quadrant for (A + B )   (I do not even understandwhat the question is asking here!!)

~                tan (A + B)      

THANK YOU FOR ANY STEP BY STEP YOU CAN HELP MEWITH.

 

Answer 1

cos(A) = 1/2 and sin(B) = -3/5.

cos(A) = 1/2 = Adjacent side / hypotenuse.

You are working with a reference angle and right triangle in quadrant IV

Opposite side = - √[(hypotenuse)² - (Adjacent side)²] = √[2² - 1²] = √[4 - 1] = √3.

sin(A) = - [Opposite side/ hypotenuse] = -√3/2

Take sin(B) = -3/5 = Opposite side / hypotenuse

You are working with a reference angle and right triangle in quadrant IV

Adjacent side = - √[(hypotenuse)² - (Opposite side)²] = √[5² - 3²] = √[25 - 9] = √16 = 4.

cos(B) =Adjacent side / hypotenuse = 4/5

Sum and Difference Formulas:

cos(A + B) = cos(A)cos(B) - sin(A)sin(B)

cos(A + B) = (1/2)(4/5) - (-√3/2)(-3/5) = (4/10) - (3√3/10)

Rewrite the expression with common denominator.

cos(A + B) = [4 - 3√3]/10

Sum and Difference Formulas:

tan((A + B) = [tan(A) + tan(B)] / [1 - tan(A)tan(B)]

tan(A) = sin(A)/cos(A) = -(√3/2)(1/2) = - √3

tan(B) = sin(B)/cos(B) = (-3/5)/(4/5) = -3/4.

[tan(A) + tan(B)] = - √3 - 3/4 = -[4√3 + 3]/4.

1 - tan(A)tan(B) = 1 - (-√3)(-3/4) = 1 - 3√3/4 = [4 - 3√3]/4

tan((A + B) = [tan(A) + tan(B)] / [1 - tan(A)tan(B)]

= {-[4√3 + 3]/4}/{[4 - 3√3]/4} = -[4√3 + 3]/[4 - 3√3]

Therefore cos(A + B) = [4 - 3√3]/10 and tan((A + B) = -[4√3 + 3]/[4 - 3√3].