Chapter T, Problem 6DDT

If $\sin x=\frac{1}{3}$ and $\sec y=\frac{5}{4}$, where x and y lie between 0 and π/2, evaluate sin(x + y).

To determine

To evaluate: The value of $\sin \left(x+y\right)$.

Answer to Problem 6DDT

The value of $\sin \left(x+y\right)$ is $\frac{1}{15}\left(4+6\sqrt{2}\right)$.

Explanation of Solution

Given:

The values are $\sin x=\frac{1}{3}\text{and}\sec y=\frac{5}{4}$, where x and y lie between 0 and $\frac{\pi }{2}$.

Formula used:

1. 1. ${\sin }^{2}x+{\cos }^{2}x=1$
2. 2. $\sin (x+y)=\sin x\cos y+\cos x\sin y$

Calculation:

Consider the value $\sin x=\frac{1}{3}$.

Use the above mentioned formula 1 and find the value of $\cos x$ as follows,

Substitute $\sin x=\frac{1}{3}$ in $\cos x=\sqrt{1-{\sin }^{2}x}$,

$\begin{array}{c}\cos x=\sqrt{1-{\left(\frac{1}{3}\right)}^2}\\ =\sqrt{1-\frac{1}{9}}\\ =\sqrt{\frac{8}{9}}\\ =\frac{2\sqrt{2}}{3}\end{array}$

Consider the value $\sec y=\frac{5}{4}$.

That is, $\cos y=\frac{4}{5}$

Use the above mentioned formula 1 and find the value of $\sin y$ as follows,

Substitute $\cos y=\frac{4}{5}$ in $\sin y=\sqrt{1-{\cos }^{2}y}$,

$\begin{array}{c}\sin y=\sqrt{1-{\left(\frac{4}{5}\right)}^2}\\ =\sqrt{1-\frac{16}{25}}\\ =\sqrt{\frac{9}{25}}\\ =\frac{3}{5}\end{array}$

Use the above mentioned formula 2 and find the value of $\sin \left(x+y\right)$ as follows,

Substitute $\cos x=\frac{2\sqrt{2}}{3}\text{and}\sin y=\frac{3}{5}$ in $\sin (x+y)=\sin x\cos y+\cos x\sin y$,

$\begin{array}{c}\sin (x+y)=\frac{1}{3}\times \frac{4}{5}+\frac{2\sqrt{2}}{3}\times \frac{3}{5}\\ =\frac{4}{15}+\frac{6\sqrt{2}}{15}\\ =\frac{1}{15}\left(4+6\sqrt{2}\right)\end{array}$

Thus, the value of $\sin \left(x+y\right)$ is $\frac{1}{15}\left(4+6\sqrt{2}\right)$.