Answer :
We start with the given equation
[tex]$$
\sec y = \frac{13}{15}.
$$[/tex]
Since the secant function is the reciprocal of the cosine function, we have
[tex]$$
\cos y = \frac{1}{\sec y} = \frac{15}{13}.
$$[/tex]
Notice that
[tex]$$
\frac{15}{13} > 1,
$$[/tex]
which is not possible for real angles because the cosine of a real angle must satisfy
[tex]$$
-1 \le \cos y \le 1.
$$[/tex]
This implies that there is no real angle [tex]\( y \)[/tex] with [tex]\( \cos y = \frac{15}{13} \)[/tex]. However, if we allow [tex]\( y \)[/tex] to be a complex angle, then we can still compute [tex]\( \sin y \)[/tex] via the Pythagorean identity:
[tex]$$
\sin^2 y + \cos^2 y = 1.
$$[/tex]
Substitute [tex]\( \cos y = \frac{15}{13} \)[/tex]:
[tex]$$
\sin^2 y = 1 - \cos^2 y = 1 - \left(\frac{15}{13}\right)^2 = 1 - \frac{225}{169}.
$$[/tex]
Write [tex]\( 1 \)[/tex] as [tex]\( \frac{169}{169} \)[/tex]:
[tex]$$
\sin^2 y = \frac{169 - 225}{169} = \frac{-56}{169}.
$$[/tex]
Since [tex]\( \sin^2 y \)[/tex] is negative, [tex]\( \sin y \)[/tex] must be a pure imaginary number when considering complex angles. Taking the square root of both sides gives:
[tex]$$
\sin y = \pm \sqrt{\frac{-56}{169}}.
$$[/tex]
We can separate the negative inside the square root:
[tex]$$
\sin y = \pm \sqrt{-1} \sqrt{\frac{56}{169}} = \pm i \sqrt{\frac{56}{169}}.
$$[/tex]
Since [tex]\( \sqrt{\frac{56}{169}} = \frac{\sqrt{56}}{13} \)[/tex] and [tex]\( \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14} \)[/tex], we obtain:
[tex]$$
\sin y = \pm \frac{2i\sqrt{14}}{13}.
$$[/tex]
Thus, if [tex]\( y \)[/tex] is allowed to be a complex angle, the solutions for [tex]\( \sin y \)[/tex] are:
[tex]$$
\boxed{\sin y = \pm \frac{2i\sqrt{14}}{13}}.
$$[/tex]
[tex]$$
\sec y = \frac{13}{15}.
$$[/tex]
Since the secant function is the reciprocal of the cosine function, we have
[tex]$$
\cos y = \frac{1}{\sec y} = \frac{15}{13}.
$$[/tex]
Notice that
[tex]$$
\frac{15}{13} > 1,
$$[/tex]
which is not possible for real angles because the cosine of a real angle must satisfy
[tex]$$
-1 \le \cos y \le 1.
$$[/tex]
This implies that there is no real angle [tex]\( y \)[/tex] with [tex]\( \cos y = \frac{15}{13} \)[/tex]. However, if we allow [tex]\( y \)[/tex] to be a complex angle, then we can still compute [tex]\( \sin y \)[/tex] via the Pythagorean identity:
[tex]$$
\sin^2 y + \cos^2 y = 1.
$$[/tex]
Substitute [tex]\( \cos y = \frac{15}{13} \)[/tex]:
[tex]$$
\sin^2 y = 1 - \cos^2 y = 1 - \left(\frac{15}{13}\right)^2 = 1 - \frac{225}{169}.
$$[/tex]
Write [tex]\( 1 \)[/tex] as [tex]\( \frac{169}{169} \)[/tex]:
[tex]$$
\sin^2 y = \frac{169 - 225}{169} = \frac{-56}{169}.
$$[/tex]
Since [tex]\( \sin^2 y \)[/tex] is negative, [tex]\( \sin y \)[/tex] must be a pure imaginary number when considering complex angles. Taking the square root of both sides gives:
[tex]$$
\sin y = \pm \sqrt{\frac{-56}{169}}.
$$[/tex]
We can separate the negative inside the square root:
[tex]$$
\sin y = \pm \sqrt{-1} \sqrt{\frac{56}{169}} = \pm i \sqrt{\frac{56}{169}}.
$$[/tex]
Since [tex]\( \sqrt{\frac{56}{169}} = \frac{\sqrt{56}}{13} \)[/tex] and [tex]\( \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14} \)[/tex], we obtain:
[tex]$$
\sin y = \pm \frac{2i\sqrt{14}}{13}.
$$[/tex]
Thus, if [tex]\( y \)[/tex] is allowed to be a complex angle, the solutions for [tex]\( \sin y \)[/tex] are:
[tex]$$
\boxed{\sin y = \pm \frac{2i\sqrt{14}}{13}}.
$$[/tex]
