Problem 36
Question
35-40 Find \(\sin \frac{x}{2}, \cos \frac{x}{2},\) and \(\tan \frac{x}{2}\) from the given information. \(\cos x=-\frac{4}{5}, \quad 180^{\circ} < x < 270^{\circ}\)
Step-by-Step Solution
Verified Answer
\( \sin \frac{x}{2} = \frac{3\sqrt{10}}{10}, \cos \frac{x}{2} = -\frac{\sqrt{10}}{10}, \tan \frac{x}{2} = -3 \).
1Step 1: Determine the value of \( \sin x \)
Since \( x \) is in the third quadrant, where both sine and cosine are negative, use the Pythagorean identity: \( \sin^2 x + \cos^2 x = 1 \). Substitute the given value of \( \cos x = -\frac{4}{5} \):\[ \sin^2 x + \left(-\frac{4}{5}\right)^2 = 1 \] \[ \sin^2 x + \frac{16}{25} = 1 \] \[ \sin^2 x = \frac{9}{25} \]. \( \sin x \) is negative in the third quadrant, so \( \sin x = -\frac{3}{5} \).
2Step 2: Calculate \( \sin \frac{x}{2} \) using the half-angle formula
The formula for \( \sin \frac{x}{2} \) is \( \pm \sqrt{\frac{1 - \cos x}{2}} \). Substitute \( \cos x = -\frac{4}{5} \): \[ \sin \frac{x}{2} = \pm \sqrt{\frac{1 + \frac{4}{5}}{2}} = \pm \sqrt{\frac{9}{10}} = \pm \frac{3\sqrt{10}}{10} \]. Since \( \frac{x}{2} \) should meet the criteria of \( 90^{\circ} < \frac{x}{2} < 135^{\circ} \), \( \sin \frac{x}{2} \) is positive, so \( \sin \frac{x}{2} = \frac{3\sqrt{10}}{10} \).
3Step 3: Calculate \( \cos \frac{x}{2} \) using the half-angle formula
The formula for \( \cos \frac{x}{2} \) is \( \pm \sqrt{\frac{1 + \cos x}{2}} \). Substitute \( \cos x = -\frac{4}{5} \): \[ \cos \frac{x}{2} = \pm \sqrt{\frac{1 - \frac{4}{5}}{2}} = \pm \sqrt{\frac{1}{10}} = \pm \frac{1}{\sqrt{10}} = \pm \frac{\sqrt{10}}{10} \]. Since \( \frac{x}{2} \) is in the second quadrant, where the cosine is negative, \( \cos \frac{x}{2} = -\frac{\sqrt{10}}{10} \).
4Step 4: Calculate \( \tan \frac{x}{2} \) using sine and cosine half-angle values
Use the identity \( \tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} \). Substitute the previously found values: \[ \tan \frac{x}{2} = \frac{\frac{3\sqrt{10}}{10}}{-\frac{\sqrt{10}}{10}} \]. Simplifying, \( \tan \frac{x}{2} = -3 \).
Key Concepts
Third QuadrantHalf-Angle FormulasPythagorean Identity
Third Quadrant
The unit circle can be divided into four quadrants, each representing a specific range of angles and corresponding sign values for trigonometric functions like sine, cosine, and tangent.
In the third quadrant, angles range from 180° to 270°. What's unique about this quadrant is:
In the third quadrant, angles range from 180° to 270°. What's unique about this quadrant is:
- The sine and cosine values are negative. This is because both the x-coordinate (cosine) and the y-coordinate (sine) in this quadrant are below their respective axes.
- The tangent, derived from the ratio of sine to cosine, is positive, essentially because a negative divided by a negative yields a positive result.
Half-Angle Formulas
Half-angle formulas are vital trigonometric identities that let us find the sine, cosine, and tangent of half an angle when the values of the whole angle are known. These formulas derive from the double-angle identities and can be expressed as:
- \( \sin \frac{x}{2} = \pm \sqrt{\frac{1 - \cos x}{2}} \)
- \( \cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}} \)
- \( \tan \frac{x}{2} = \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} \)
Pythagorean Identity
The Pythagorean identity is an essential building block in trigonometry, closely related to the unit circle. It states: \[ \sin^2 x + \cos^2 x = 1 \]This identity is derived from the Pythagorean theorem and highlights the relationship between the sine and cosine of an angle.
- If one of the values (\( \sin x \) or \( \cos x \)) is known, the other can be easily determined using this identity.
- It also serves as the foundation for many other trigonometric identities and equations.
Other exercises in this chapter
Problem 36
Find all solutions of the equation. $$3 \tan ^{3} x-3 \tan ^{2} x-\tan x+1=0$$
View solution Problem 36
\(23-40\) . Prove the identity. $$ 1-\tan x \tan y=\frac{\cos (x+y)}{\cos x \cos y} $$
View solution Problem 37
Verify the identity. $$ (1-\cos \beta)(1+\cos \beta)=\frac{1}{\csc ^{2} \beta} $$
View solution Problem 37
Find all solutions of the equation. $$4 \sin x \cos x+2 \sin x-2 \cos x-1=0$$
View solution