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Find the exact values of \(\sin (\theta / 2), \cos (\theta / 2),\) and \(\tan (\theta / 2)\) for the given conditions. $$\csc \theta=-\frac{5}{3} ;-90^{\circ}<\theta<0^{\circ}$$

Short Answer

Expert verified
\(\sin(\theta/2) = \frac{1}{\sqrt{10}}, \cos(\theta/2) = \frac{3}{\sqrt{10}}, \tan(\theta/2) = -\frac{1}{3}\).

Step by step solution

01

Interpret the Given Condition

We are given that \( \csc \theta = -\frac{5}{3} \) and \( -90^{\circ} < \theta < 0^{\circ} \). This means \( \theta \) is in the fourth quadrant where \( \sin \theta = -\frac{1}{\csc \theta} \).
02

Find \( \sin \theta \)

Since \( \csc \theta = -\frac{5}{3} \), we have \( \sin \theta = -\frac{1}{\csc \theta} = -\frac{1}{-\frac{5}{3}} = -\frac{3}{5} \).
03

Find \( \cos \theta \)

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we substitute \( \sin \theta = -\frac{3}{5} \): \((-\frac{3}{5})^2 + \cos^2 \theta = 1 \). \Thus, \( \frac{9}{25} + \cos^2 \theta = 1 \), so \( \cos^2 \theta = \frac{16}{25} \), giving \( \cos \theta = \pm \frac{4}{5} \). Since \( \theta \) is in the fourth quadrant, \( \cos \theta \) is positive: \( \cos \theta = \frac{4}{5} \).
04

Use Half-Angle Identities

We use half-angle identities: - \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}} \) - \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} \) - \( \tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta} \) or \( \frac{1 - \cos \theta}{\sin \theta} \).
05

Calculate \( \sin \frac{\theta}{2} \)

Given \( \cos \theta = \frac{4}{5} \), calculate \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \frac{4}{5}}{2}} = \pm \sqrt{\frac{1}{10}} = \pm \frac{1}{\sqrt{10}} \). Fortunately, \( -90^{\circ} < \theta < 0^{\circ} \) implies \( \frac{\theta}{2} \) is in the second quadrant, so \( \sin \frac{\theta}{2} > 0 \): \( \sin \frac{\theta}{2} = \frac{1}{\sqrt{10}} \).
06

Calculate \( \cos \frac{\theta}{2} \)

Using \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} = \pm \sqrt{\frac{9}{10}} = \pm \frac{3}{\sqrt{10}} \). Since \( \frac{\theta}{2} \) is in the second quadrant where \( \cos > 0 \), \( \cos \frac{\theta}{2} = \frac{3}{\sqrt{10}} \).
07

Calculate \( \tan \frac{\theta}{2} \)

Using \( \tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta} = \frac{-\frac{3}{5}}{1 + \frac{4}{5}} = \frac{-\frac{3}{5}}{\frac{9}{5}} = -\frac{1}{3} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Angle Identities
Half-angle identities are powerful tools in trigonometry that help us find the values of sine, cosine, and tangent at half of a given angle. These identities are derived from the well-known double-angle identities and are crucial when solving problems involving specific trigonometric values. Here's a closer look at each identity:
  • For sine, the half-angle identity is \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2}} \). This allows us to find the sine of half the angle when we know the cosine of the full angle.
  • For cosine, the formula is \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos \theta}{2}} \). Similarly, this takes the known value of cosine for an angle to determine the cosine for half of that angle.
  • For tangent, the identity can be expressed as \( \tan \frac{\theta}{2} = \frac{1 - \cos \theta}{\sin \theta} \) or \( \frac{\sin \theta}{1 + \cos \theta} \). This helps find the tangent of half an angle in terms of sine and cosine of the original angle.
Choosing the correct sign for these expressions depends heavily on the quadrant where the half-angle lies, reinforcing the importance of understanding quadrants in trigonometry.
Quadrants of the Unit Circle
The unit circle is a circle with a radius of one centered at the origin of a coordinate plane. It is divided into four quadrants, each having specific properties that determine the signs of trigonometric functions:
  • Quadrant I: Ranges from \(0^\circ\) to \(90^\circ\). In this quadrant, both sine and cosine are positive.
  • Quadrant II: From \(90^\circ\) to \(180^\circ\), sine remains positive while cosine becomes negative.
  • Quadrant III: Extending from \(180^\circ\) to \(270^\circ\), both sine and cosine are negative.
  • Quadrant IV: Ranging from \(270^\circ\) to \(360^\circ\), sine is negative while cosine returns to positive.
In this exercise, since \(-90^{\circ} < \theta < 0^{\circ}\), \(\theta\) is positioned in Quadrant IV. Hence, sine is negative and cosine remains positive. This quadrant principle helps determine the signs of resultants for identities like the half-angle identities, as seen in determining the sign of \(\sin \frac{\theta}{2}\) and \(\cos \frac{\theta}{2}\). Understanding these quadrants supports accurate application of trigonometric identities.
Trigonometric Functions
Trigonometric functions are fundamental in mathematics, describing the relationship between angles and sides in right-angled triangles. The three primary functions—sine, cosine, and tangent—are often derived or utilized through identities, like in this problem.
  • **Sine** \(\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}\): Indicates the ratio of the side opposite an angle to the hypotenuse in a right triangle.
  • **Cosine** \(\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}}\): Represents the ratio of the adjacent side to the hypotenuse.
  • **Tangent** \(\tan \theta = \frac{\sin \theta}{\cos \theta}\): Exposes the ratio of the opposite to the adjacent side, or effectively, sine over cosine.
In this context, we begin with knowing the cosecant \( \csc \theta = -\frac{5}{3} \), which gives us the sine as \( \sin \theta = -\frac{3}{5} \), since cosecant is the reciprocal of sine. Next, using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we find cosine. These foundational relationships establish the necessary values, enabling the accurate use of half-angle identities and determination of the exact trigonometric values at half-angles.

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