Notes: `sec(v - u)` Find the exact value of the trigonometric expression given that sin(u) = 5/13 and cos(v) = -3/5 (both u and v are in quadrant II.)

Saturday, 11 June 2016

$\displaystyle{\sec{{\left({v}-{u}\right)}}}$ Find the exact value of the trigonometric expression given that sin(u) = 5/13 and cos(v) = -3/5 (both u and v are in quadrant II.)

There are a lot of ways to solve this problem. For instance, one can use the secant sum/difference identity. This identity, however, has a lot of terms. A more elegant solution would be to use the cosine sum/difference identity, and relating cosine to secant.

Note that $\displaystyle{\sec{{\left({x}\right)}}}=\frac{{1}}{{{\cos{{\left({x}\right)}}}}}$ .Such that we only need to get the reciprocal of the value of cosine at our given angle to get the corresponding secant. Both u and v are in the second quadrant implying that the x-components are both negative and the y-components positive.

The cosine sum/difference identity is: $\displaystyle{\cos{{\left({v}-{u}\right)}}}={\cos{{\left({v}\right)}}}{\cos{{\left({u}\right)}}}+{\sin{{\left({v}\right)}}}{\sin{{\left({u}\right)}}}$ .

We already know some of the terms here, namely: $\displaystyle{\sin{{\left({u}\right)}}}=\frac{{5}}{{13}}$ and $\displaystyle{\cos{{\left({v}\right)}}}=-\frac{{3}}{{5}}$ .To get the remaining terms, we simply use the Pythegorean identity: $\displaystyle{{\sin}}^{{2}}{\left({x}\right)}+{{\cos}}^{{2}}{\left({x}\right)}={1}$ .

Hence:

$\displaystyle{{\cos}}^{{2}}{\left({u}\right)}={1}-\frac{{25}}{{169}}=\frac{{144}}{{169}}$ , and $\displaystyle{\cos{{\left({u}\right)}}}=-\frac{{12}}{{13}}$ . Note that we could have chosen the positive root, but since u is in the second quadrant, we get the negative one.

Similarly:

$\displaystyle{{\sin}}^{{2}}{\left({v}\right)}={1}-\frac{{9}}{{25}}=\frac{{16}}{{25}}$ , and $\displaystyle{\sin{{\left({v}\right)}}}=\frac{{4}}{{5}}$ . We are getting the positive root since v is in the second quadrant.

Now, we can solve for the cosine:

$\displaystyle{\cos{{\left({v}-{u}\right)}}}={\left(-\frac{{3}}{{5}}\right)}{\left(-\frac{{12}}{{13}}\right)}+{\left(\frac{{4}}{{5}}\right)}{\left(\frac{{5}}{{13}}\right)}=\frac{{36}}{{65}}+\frac{{20}}{{65}}=\frac{{56}}{{65}}.$