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For positive acute angles AA and B,B, it is known that \sin A=\frac{8}{17}sinA= 17 8 ​ and \cos B=\frac{24}{25}. CosB= 25 24 ​. Find the value of \sin(A+B)sin(A+B) in simplest form

Answer :

Sqdancefango

Answer:

  sin(A+B) = 297/425

Step-by-step explanation:

This problem involves a couple of Pythagorean triples, and the trig identity for the sine of the sum of angles.

The Pythagorean triples involved are {8, 15, 17} and {7, 24, 25}. These tell us the other trig functions of the given angles:

  sin(A) = 8/17   ⇔   cos(A) = 15/17

  cos(B) = 24/25   ⇔   sin(B) = 7/25

The trig identity we're using is ...

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

  sin(A+B) = (8/17)(24/25) +(15/17)(7/25)

  [tex]\sin(A+B)=\dfrac{8\cdot24+15\cdot 7}{17\cdot25}\\\\\boxed{\sin(A+B)=\dfrac{297}{425}}[/tex]

_____

Check

  A = arcsin(8/17) ≈ 28.072°

  B = arccos(24/25) ≈ 16.260°

  sin(A+B) = sin(44.332°) ≈ 0.6988235

  sin(A+B) = 297/425

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