Respuesta :
We know that angles a and b are in the first quadrant. We also know this values:
[tex]\begin{gathered} \sin a=\frac{5}{13} \\ \cos b=\frac{3}{5} \end{gathered}[/tex]We have to find sin(a+b).
We can use the following identity:
[tex]\sin (a+b)=\sin a\cdot\cos b+\cos a\cdot\sin b[/tex]For the second term, we can replace the factors with another identity:
[tex]\sin (a+b)=\sin a\cdot\cos b+\sqrt[]{1-\sin^2a}\cdot\sqrt[]{1-\cos^2b}[/tex]Now we know all the terms from the right side of the equation and we can calculate:
[tex]\begin{gathered} \sin (a+b)=\sin a\cdot\cos b+\sqrt[]{1-\sin^2a}\cdot\sqrt[]{1-\cos^2b} \\ \sin (a+b)=\frac{5}{13}\cdot\frac{3}{5}+\sqrt[]{1-(\frac{5}{13})^2}\cdot\sqrt[]{1-(\frac{3}{5})^2} \\ \sin (a+b)=\frac{15}{65}+\sqrt[]{1-\frac{25}{169}}\cdot\sqrt[]{1-\frac{9}{25}} \\ \sin (a+b)=\frac{15}{65}+\sqrt[]{\frac{169-25}{169}}\cdot\sqrt[]{\frac{25-9}{25}} \\ \sin (a+b)=\frac{15}{65}+\sqrt[]{\frac{144}{169}}\cdot\sqrt[]{\frac{16}{25}} \\ \sin (a+b)=\frac{15}{65}+\frac{12}{13}\cdot\frac{4}{5} \\ \sin (a+b)=\frac{15}{65}+\frac{48}{65} \\ \sin (a+b)=\frac{63}{65} \end{gathered}[/tex]Answer: sin(a+b) = 63/65
