Respuesta :
The trigonometric identity sin(x+y) is given by the following formula:
[tex]\sin (x+y)=\sin x\cdot\cos y+\cos x\cdot\sin y[/tex]We need to find then cos(y), cos(x) and sin(y).
Other trigonometric identities we can use are:
[tex]\begin{gathered} \sec y=\frac{1}{\cos y}\text{ Identity (1)} \\ \sin ^2a+\cos ^2a=1\text{ Identity (2)} \end{gathered}[/tex]First, replace the sec(y) value and find cos(y):
[tex]\begin{gathered} \frac{5}{3}=\frac{1}{\cos y} \\ \therefore\cos y=\frac{1\times3}{5}=\frac{3}{5} \end{gathered}[/tex]Now, apply the second identity to find cos(x) and sin(y):
[tex]\begin{gathered} \sin ^2x+\cos ^2x=1 \\ \therefore\cos ^2x=1-\sin ^2x \\ \therefore\sqrt[]{\cos^2x}=\sqrt[]{1-\sin^2x} \\ \therefore\cos x=\sqrt[]{1-(\frac{1}{4})^2} \\ \cos x=\sqrt[]{1-\frac{1}{16}^{}}=\sqrt[]{\frac{1\cdot16-1\cdot1}{16}} \\ \cos x=\sqrt[]{\frac{15}{16}}=\frac{\sqrt[]{15}}{\sqrt[]{16}}=\frac{\sqrt[]{15}}{4} \end{gathered}[/tex]And now sin y=
[tex]\begin{gathered} \sin ^2y=1-\cos ^2y \\ \therefore\sin y=\sqrt[]{1-\cos ^2y} \\ \therefore\sin y=\sqrt[]{1-(\frac{3}{5})^2} \\ \sin y=\sqrt[]{1-\frac{9}{25}}=\sqrt[]{\frac{1\cdot25-1\cdot9}{25}} \\ \sin y=\sqrt[]{\frac{16}{25}}=\frac{\sqrt[]{16}}{\sqrt[]{25}}=\frac{4}{5} \end{gathered}[/tex]Now, replace these values into the identity sin(x+y) and solve as follows:
[tex]\begin{gathered} \sin (x+y)=\sin x\cdot\cos y+\cos x\cdot\sin y \\ \sin (x+y)=\frac{1}{4}\cdot\frac{3}{5}+\frac{\sqrt[]{15}}{4}\cdot\frac{4}{5} \\ \sin (x+y)=\frac{1\times3}{4\times5}+\frac{\sqrt[]{15}\times4}{4\times5} \\ \sin (x+y)=\frac{3}{20}+\frac{4\sqrt[]{15}}{20}=\frac{3+4\sqrt[]{15}}{20} \\ \sin (x+y)\approx0.925 \end{gathered}[/tex]