Question

Angle \(\theta\) is in quadrant II and \(\sin \theta=\frac{5}{13} .\) Determine an exact value for each of the following. a) \(\cos 2 \theta\) b) \(\sin 2 \theta\) c) \(\sin \left(\theta+\frac{\pi}{2}\right)\)

Short answer

a) \frac{119}{169} b) -\frac{120}{169} c) -\frac{12}{13}

Step-by-step solution

Determine \cos \theta

Use the Pythagorean identity \sin^2 \theta + \cos^2 \theta = 1 to find \cos \theta. Given \sin \theta = \frac{5}{13}, we have \(\sin^2 \theta = \(\frac{5}{13}\)^2 = \frac{25}{169}\). Then \cos^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}. Since \theta is in quadrant II, \cos \theta is negative: \cos \theta = -\sqrt{\frac{144}{169}} = -\frac{12}{13}.

Compute \cos 2\theta

Use the double angle formula for cosine: \cos 2\theta = \cos^2 \theta - \sin^2 \theta. Substitute \cos \theta = -\frac{12}{13} and \sin \theta = \frac{5}{13}. \[ \cos 2\theta = \left(\frac{12}{13}\right)^2 - \left(\frac{5}{13}\right)^2 = \frac{144}{169} - \frac{25}{169} = \frac{119}{169}\]\.

Compute \sin 2\theta

Use the double angle formula for sine: \sin 2\theta = 2 \sin \theta \cos \theta. Substitute \sin \theta = \frac{5}{13} and \cos \theta = -\frac{12}{13}. \[ \sin 2\theta = 2\left(\frac{5}{13}\right)\left(-\frac{12}{13}\right) = -2\frac{60}{169} = -\frac{120}{169} = -\frac{120}{169}\]\.

Compute \sin \(\theta + \frac{\pi}{2}\)

Use the sine addition formula \sin(a + b): \sin(a + b) = \sin a \cos b + \cos a \sin b. For \(a = \theta\) and \(b = \frac{\pi}{2}\): \sin \(\theta + \frac{\pi}{2}\) = \sin \theta \cos \(\frac{\pi}{2}\) + \cos \theta \sin \(\frac{\pi}{2}\). Since \cos \(\frac{\pi}{2}\) = 0 and \sin \(\frac{\pi}{2}\) = 1, this simplifies to: \[ \sin(\theta + \frac{\pi}{2}) = (\frac{5}{13})(0) + (-\frac{12}{13})(1) = -\frac{12}{13}\]\.

Key concepts

Pythagorean Identity

The Pythagorean identity is a fundamental concept in trigonometry. It is given by \(\sin^2 \theta + \cos^2 \theta = 1\). Basically, this identity helps us to find one trigonometric function if we know the other. In this exercise, we know that \sin \theta = \frac{5}{13}\. So, we can substitute this into our identity.
\(\sin^2 \theta = \left(\frac{5}{13}\right)^2 = \frac{25}{169}\)\.
Next, we solve for \(\cos^2 \theta\) using the identity.
\(\cos^2 \theta = 1 - \sin^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}\)\.
Knowing the quadrant is key. Since \theta\ is in quadrant II, \cos \theta\ will be negative. Thus, \(\cos \theta = -\sqrt{\frac{144}{169}} = -\frac{12}{13}\)\.
The Pythagorean identity is very useful for solving trigonometric problems.

Double Angle Formulas

Double angle formulas allow us to express trigonometric functions of double angles (2\theta) in terms of single angles (\theta). The formulas are:
\(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\)\ and \(\sin 2\theta = 2 \sin \theta \cos \theta\)\.
Let's substitute the values we found earlier:
For \(\cos 2 \theta\), we have:
\(\cos 2 \theta = \left(\frac{12}{13}\right)^2 - \left(\frac{5}{13}\right)^2 = \frac{144}{169} - \frac{25}{169} = \frac{119}{169}\)\.
And for \(\sin 2 \theta\), we use:
\(\sin 2 \theta = 2 \left(\frac{5}{13}\right) \left(-\frac{12}{13}\right) = -2\frac{60}{169} = -\frac{120}{169}\)\.
These formulas are very handy for transforming and solving trigonometric equations involving double angles.

Sine Addition Formula

The sine addition formula helps to find the sine of the sum of two angles. The formula is:
\(\sin(a + b) = \sin a \cos b + \cos a \sin b\)\. In this exercise, we are asked to compute \(\sin\left( \theta + \frac{\pi}{2} \right)\)\.
So, we set \(\theta = a\) and \(\frac{\pi}{2} = b\). Substituting into our sine addition formula gives:
\(\sin(\theta + \frac{\pi}{2}) = \sin \theta \cos(\frac{\pi}{2}) + \cos \theta \sin(\frac{\pi}{2})\)\.
We know that \(\cos\left(\frac{\pi}{2}\right) = 0\) and \(\sin\left(\frac{\pi}{2}\right) = 1\). Substituting these values, we get:
\(\sin(\theta + \frac{\pi}{2}) = (\frac{5}{13})(0) + (\-\frac{12}{13})(1) = -\frac{12}{13}\)\.
The sine addition formula makes it easy to manage sums of angles in trigonometric functions.