Angle Sum and Difference

Question

For positive acute angles A and B, it is known that sinA= 21/29 and cosB= 7/25. Find the value of sin(A+B) in simplest form.

Answer 1

when the angle is positive and acute it's in quadrant I

0<A<pi/2 and 0<B<pi/2

sineA = 21/29 = opposite side over hypotenuse = o/h

adjacent side = square root of hypotenuse squared minus opposite side squared

= sqr(h^2 - o^2) = sqr(29^2 - 21^2) = sqr(841-441)=sqr400 = 20

cosA = 20/29

cosB = 7/25 = a/h. opposite side = o= sqr(h^2-a^2) = sqr(25^2-7^2) = sqr(625-

49)= sqr576 = 24

sinB = o/h = 24/25

sin(A+B) = sinAcosB+sinBcosA = (21/29)(7/25)+(24/25)(20/29) =

(147+480)/(29)(25) = 627/725= about 0.865

627/725 is an exact solution

0.865 is an approximation rounded to 3 decimal places

check the answer by directly finding A and B, add them and take the sine

with a calculator using a trig function

cosB=7/25 B=cos^-1(7/25) = 73.74 degrees

sinA =21/29 A = sin^-1(21/29) = 46.40

A+B = 73.74 +46.40 = 120.14

sin120.14 = 0.865

A+B is not acute, but sin120.14 is positive, in quadrant II

when you get the same solution, working the problem 2 different ways, you can feel very confident you got the right answer

Answer 2

sin(A+B) = sinAcosB+sinBcosA

if cosB = 7/25 then sinB =24/25 It's a 7-24-25 right triangle

if sinA =21/29 then cosA =20/29 a 20-21-29 right triangle

plug those values into 1st equation above

sin(21/29)cos(7/25)+sin(24/25)cos(20/29) =1.268667562