Ex 3: Cos2x = -6 where 2x is in 15 the second quadrant. Find the EXACT valu of cosx. V Cos2x = -6 صاف يا اب 2 cos²x-1=-6 2 cos²x = -6 +1 o/w /w/outo úlo 15 2 cos²x = -6 + 15 2 cos²x = 9 15 15 cos²x = 3 2 616 5-2 cos²x = 3 - 30 " I 42x4 TT 2 :) d TLX21 2 =) x is in the first quadrant. :: COS X = +√√30 = +√3 lo UTO
Ex 3: Cos2x = -6 where 2x is in 15 the second quadrant. Find the EXACT valu of cosx. V Cos2x = -6 صاف يا اب 2 cos²x-1=-6 2 cos²x = -6 +1 o/w /w/outo úlo 15 2 cos²x = -6 + 15 2 cos²x = 9 15 15 cos²x = 3 2 616 5-2 cos²x = 3 - 30 " I 42x4 TT 2 :) d TLX21 2 =) x is in the first quadrant. :: COS X = +√√30 = +√3 lo UTO
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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![## Example 3: Solving Cosine Equation
Given:
\[
\cos 2x = -\frac{6}{15}
\]
where \(2x\) is in the second quadrant. Find the **exact value** of \(\cos x\).
### Solution:
1. Start with the equation:
\[
\cos 2x = -\frac{6}{15}
\]
2. Use the double angle identity:
\[
2\cos^2 x - 1 = -\frac{6}{15}
\]
3. Rearrange and solve for \(\cos^2 x\):
\[
2\cos^2 x = -\frac{6}{15} + 1
\]
\[
2\cos^2 x = \frac{-6 + 15}{15}
\]
\[
2\cos^2 x = \frac{9}{15}
\]
\[
\cos^2 x = \frac{9}{15 \times 2} = \frac{3}{10}
\]
4. Solve for \(\cos x\):
\[
\cos x = \pm \sqrt{\frac{3}{10}}
\]
\[
\cos x = \pm \frac{\sqrt{3}}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}} = \pm \frac{\sqrt{30}}{10}
\]
5. Determine the sign of \(\cos x\):
- Since \(\frac{\pi}{4} < x < \frac{\pi}{2}\), \(x\) is in the **first quadrant**.
- In the first quadrant, cosine is positive.
Thus, \(\cos x = +\frac{\sqrt{30}}{10} = +\frac{\sqrt{3}}{\sqrt{10}}\).
### Diagram Explanation
The diagram in the solution depicts the trigonometric unit circle showing angle \(2x\) in the second quadrant. The quadrant criteria indicate that the value of \(\cos x\) will be positive in the first quadrant, consistent with the calculation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd298be25-ffcb-47a1-b62e-d3a828be4885%2Fad2da373-d563-4987-b42f-8aa93df52ef8%2Fiz0mqq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Example 3: Solving Cosine Equation
Given:
\[
\cos 2x = -\frac{6}{15}
\]
where \(2x\) is in the second quadrant. Find the **exact value** of \(\cos x\).
### Solution:
1. Start with the equation:
\[
\cos 2x = -\frac{6}{15}
\]
2. Use the double angle identity:
\[
2\cos^2 x - 1 = -\frac{6}{15}
\]
3. Rearrange and solve for \(\cos^2 x\):
\[
2\cos^2 x = -\frac{6}{15} + 1
\]
\[
2\cos^2 x = \frac{-6 + 15}{15}
\]
\[
2\cos^2 x = \frac{9}{15}
\]
\[
\cos^2 x = \frac{9}{15 \times 2} = \frac{3}{10}
\]
4. Solve for \(\cos x\):
\[
\cos x = \pm \sqrt{\frac{3}{10}}
\]
\[
\cos x = \pm \frac{\sqrt{3}}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}} = \pm \frac{\sqrt{30}}{10}
\]
5. Determine the sign of \(\cos x\):
- Since \(\frac{\pi}{4} < x < \frac{\pi}{2}\), \(x\) is in the **first quadrant**.
- In the first quadrant, cosine is positive.
Thus, \(\cos x = +\frac{\sqrt{30}}{10} = +\frac{\sqrt{3}}{\sqrt{10}}\).
### Diagram Explanation
The diagram in the solution depicts the trigonometric unit circle showing angle \(2x\) in the second quadrant. The quadrant criteria indicate that the value of \(\cos x\) will be positive in the first quadrant, consistent with the calculation.
![### Trigonometric Identities and Calculations
#### Part b: Simplification of Trigonometric Expression
Given the expression:
\[ \cos^2\left(\frac{7\pi}{8}\right) - \sin^2\left(\frac{7\pi}{8}\right) \]
1. Using the trigonometric identity:
\[ \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \]
2. Substitute \(\theta = \frac{7\pi}{16}\):
\[ \cos\left(2 \times \frac{7\pi}{16}\right) = \cos\left(\frac{7\pi}{8}\right) \]
3. Simplified to:
\[ = \cos\left(\frac{7\pi}{4}\right) \]
4. Which equals:
\[ = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \]
5. Final result:
\[ = \frac{\sqrt{2}}{2} \]
A unit circle diagram accompanies this solution, showing a right triangle with angles and a labeled hypotenuse of 1, illustrating the calculation geometrically.
#### Example 2: Finding \(\cos 2\theta\)
Given:
\[ \tan\theta = -\frac{3}{4} \]
\[ \text{where } \frac{3\pi}{2} \leq \theta \leq 2\pi \]
To find \(\cos 2\theta\):
1. Use the identity for cosine double angle:
\[ \cos 2\theta = 2\cos^2\theta - 1 \]
2. Calculate cosine using the right triangle in the unit circle where angle \(\theta\) is in the fourth quadrant:
- Hypotenuse (r): 5
- Opposite side: -3
- Adjacent side: 4
3. Calculate:
\[ \cos\theta = \frac{4}{5} \]
4. Substitute into the identity:
\[ \cos 2\theta = 2\left(\frac{4}{5}\right)^2 - 1 \]
5. Simplify:
\[ = 2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd298be25-ffcb-47a1-b62e-d3a828be4885%2Fad2da373-d563-4987-b42f-8aa93df52ef8%2F7p1u4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Trigonometric Identities and Calculations
#### Part b: Simplification of Trigonometric Expression
Given the expression:
\[ \cos^2\left(\frac{7\pi}{8}\right) - \sin^2\left(\frac{7\pi}{8}\right) \]
1. Using the trigonometric identity:
\[ \cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) \]
2. Substitute \(\theta = \frac{7\pi}{16}\):
\[ \cos\left(2 \times \frac{7\pi}{16}\right) = \cos\left(\frac{7\pi}{8}\right) \]
3. Simplified to:
\[ = \cos\left(\frac{7\pi}{4}\right) \]
4. Which equals:
\[ = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} \]
5. Final result:
\[ = \frac{\sqrt{2}}{2} \]
A unit circle diagram accompanies this solution, showing a right triangle with angles and a labeled hypotenuse of 1, illustrating the calculation geometrically.
#### Example 2: Finding \(\cos 2\theta\)
Given:
\[ \tan\theta = -\frac{3}{4} \]
\[ \text{where } \frac{3\pi}{2} \leq \theta \leq 2\pi \]
To find \(\cos 2\theta\):
1. Use the identity for cosine double angle:
\[ \cos 2\theta = 2\cos^2\theta - 1 \]
2. Calculate cosine using the right triangle in the unit circle where angle \(\theta\) is in the fourth quadrant:
- Hypotenuse (r): 5
- Opposite side: -3
- Adjacent side: 4
3. Calculate:
\[ \cos\theta = \frac{4}{5} \]
4. Substitute into the identity:
\[ \cos 2\theta = 2\left(\frac{4}{5}\right)^2 - 1 \]
5. Simplify:
\[ = 2
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