need help please, college math

Question

Find the EXACT value of  if sin A =  and cos B = . Assume A and B are Quadrant I angles in standard position. = 

Answer 1

From trigonometric identity for angle subtraction;

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

Since, sin A = 3/10 and cos B = √(13)/7; solve for cos A and sin B.

Using the Pythagorean identity;

sin² A + cos² A = 1 (1) {For angle A}

sin² B + cos² B = 1 (2) {For angle B}

From equation (1), solving for cos A;

cos² A = 1 - sin² A

cos A = √(1 - sin² A)

Then,

cos A = √[1 - (3/10)²] = √[1 - (9/100)] = √(91/100) = √(91)/10

From equation (2), solving for sin B;

sin² B = 1 - cos² B

sin B = √(1 - cos² B)

Then,

sin B = √[1 - (√(13)/7)²) = √[1 - (13/49)] = √(36/49) = 6/7

Finally,

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

= (3/10)*[√(13)/7] - [√(91)/10]*(6/7)

= [3√(13)]/70 - [6√(91)]/70

= (1/70)*[3√(13) - 6√(91)] NB: √(91) = √(13)*√(7)

= (1/70)*{3√(13) - 3√(13)*[2√(7)]}

sin(A - B) = [3√(13)/70]*[1 - 2√(7)] ~= -0.6631 (to 4 significant figures)

Answer 2

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

sin (A - B) = (3/10)(√13/7) - (√98/10)(4/7)

Can you simplify and answer?