Answer: -16/65
Step-by-step explanation:
Drawing the right triangle (as attached) gives us that [tex]\arctan \left(\frac{12}{5} \right)=\arcsin \left(\frac{12}{13} \right)[/tex]
Also, [tex]-\arcsin \left(-\frac{3}{5} \right)=\arcsin \left(\frac{3}{5} \right)[/tex]
This means our original expression is equal to:
[tex]\cos \left[\arcsin \left(\frac{12}{13} \right)+\arcsin \left(\frac{3}{5} \right) \right][/tex]
Using the cosine addition formula, which states [tex]\cos(a+b)=\cos a \cos b-\sin a \sin b[/tex], we get this itself is equal to:
[tex]\cos \left(\arcsin \left(\frac{12}{13} \right) \right)\cos \left(\arcsin \left(\frac{3}{5} \right)\right)-\sin \left(\arcsin \left(\frac{12}{13} \right) \right)\sin \left(\arcsin \left(\frac{3}{5} \right)\right)[/tex]
Since [tex]\sin^{2} \theta+\cos^{2} \theta=1[/tex], we know that:
[tex]\sin^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)+\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\\frac{144}{169} +\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=1\\\\cos^{2} \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{25}{169}\\\\cos \left(\arcsin \left(\frac{12}{13} \right)\right)=\frac{5}{13}[/tex]
Similarly, cos(arcsin(3/5))=4/5.
This means the given expression is equal to:
[tex]\left(\frac{5}{13} \right) \left(\frac{4}{5} \right)-\left(\frac{12}{13} \right) \left(\frac{3}{5} \right)\\\\\frac{20}{65}-\frac{36}{65}=\boxed{-\frac{16}{65}}[/tex]
