Answer: 312/1537
Explanation
We'll need the Pythagorean trig identity
[tex]\sin^2( \text{x} ) + \cos^2( \text{x} ) = 1[/tex]
If sin(A) = 45/53, then it leads to cos(A) = 28/53 because of this scratch work shown here:
[tex]\sin^2( A ) + \cos^2( A ) = 1\\\\(45/53)^2 + \cos^2( A ) = 1\\\\2025/2809 + \cos^2( A ) = 1\\\\\cos^2( A ) = 1-2025/2809\\\\[/tex]
[tex]\cos^2( A ) = 784/2809\\\\\cos( A ) = \sqrt{784/2809}\\\\\cos( A ) = \sqrt{784}/\sqrt{2809}\\\\\cos( A ) = 28/53\\\\[/tex]
Since angle A is acute, it means cos(A) is positive.
Through similar reasoning and similar scratch work, cos(B) = 20/29 leads to sin(B) = 21/29.
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The key takeaway from the last section is the following four items
- sin(A) = 45/53
- cos(A) = 28/53
- cos(B) = 20/29
- sin(B) = 21/29
Those items are plugged into the identity shown here
sin(A-B) = sin(A)cos(B) - cos(A)sin(B)
sin(A-B) = (45/53)*(20/29) - (28/53)*(21/29)
sin(A-B) = 312/1537
