Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
To find the values of [tex]$\sin \theta$[/tex] and [tex]$\cos \theta$[/tex] given that [tex]$\tan \theta = -\frac{3}{4}$[/tex] and [tex]$\cos \theta < 0$[/tex], we can follow these steps:
1. Understand the given information:
- We know that [tex]$\tan \theta = \frac{\sin \theta}{\cos \theta}$[/tex].
- Therefore, we are given that [tex]$\frac{\sin \theta}{\cos \theta} = -\frac{3}{4}$[/tex].
2. Express [tex]$\sin \theta$[/tex] and [tex]$\cos \theta$[/tex] in terms of a common variable:
- Let [tex]$\sin \theta = -3k$[/tex] (since the ratio of [tex]$\sin \theta$[/tex] to [tex]$\cos \theta$[/tex] is [tex]$-\frac{3}{4}$[/tex]).
- Let [tex]$\cos \theta = 4k$[/tex].
3. Use the Pythagorean identity:
- Recall that for any angle [tex]$\theta$[/tex], [tex]$\sin^2 \theta + \cos^2 \theta = 1$[/tex].
- Substituting our expressions for [tex]$\sin \theta$[/tex] and [tex]$\cos \theta$[/tex], we get:
[tex]\[ (-3k)^2 + (4k)^2 = 1 \][/tex]
Simplify the equation:
[tex]\[ 9k^2 + 16k^2 = 1 \][/tex]
[tex]\[ 25k^2 = 1 \][/tex]
[tex]\[ k^2 = \frac{1}{25} \][/tex]
[tex]\[ k = \pm \frac{1}{5} \][/tex]
4. Determine the correct value of [tex]\( k \)[/tex]:
- Since [tex]$\cos \theta < 0$[/tex], and [tex]$\cos \theta = 4k$[/tex], [tex]\( k \)[/tex] must be negative.
- Therefore, [tex]\( k = -\frac{1}{5} \)[/tex].
5. Find the values of [tex]$\sin \theta$[/tex] and [tex]$\cos \theta$[/tex]:
- Substitute [tex]\( k \)[/tex] back into our expressions for [tex]$\sin \theta$[/tex] and [tex]$\cos \theta$[/tex]:
[tex]\[ \sin \theta = -3k = -3 \left(-\frac{1}{5}\right) = \frac{3}{5} = 0.6 \][/tex]
[tex]\[ \cos \theta = 4k = 4 \left(-\frac{1}{5}\right) = -\frac{4}{5} = -0.8 \][/tex]
Therefore, the values of the trigonometric functions are:
[tex]\[ \sin \theta = 0.6 \][/tex]
[tex]\[ \cos \theta = -0.8 \][/tex]
1. Understand the given information:
- We know that [tex]$\tan \theta = \frac{\sin \theta}{\cos \theta}$[/tex].
- Therefore, we are given that [tex]$\frac{\sin \theta}{\cos \theta} = -\frac{3}{4}$[/tex].
2. Express [tex]$\sin \theta$[/tex] and [tex]$\cos \theta$[/tex] in terms of a common variable:
- Let [tex]$\sin \theta = -3k$[/tex] (since the ratio of [tex]$\sin \theta$[/tex] to [tex]$\cos \theta$[/tex] is [tex]$-\frac{3}{4}$[/tex]).
- Let [tex]$\cos \theta = 4k$[/tex].
3. Use the Pythagorean identity:
- Recall that for any angle [tex]$\theta$[/tex], [tex]$\sin^2 \theta + \cos^2 \theta = 1$[/tex].
- Substituting our expressions for [tex]$\sin \theta$[/tex] and [tex]$\cos \theta$[/tex], we get:
[tex]\[ (-3k)^2 + (4k)^2 = 1 \][/tex]
Simplify the equation:
[tex]\[ 9k^2 + 16k^2 = 1 \][/tex]
[tex]\[ 25k^2 = 1 \][/tex]
[tex]\[ k^2 = \frac{1}{25} \][/tex]
[tex]\[ k = \pm \frac{1}{5} \][/tex]
4. Determine the correct value of [tex]\( k \)[/tex]:
- Since [tex]$\cos \theta < 0$[/tex], and [tex]$\cos \theta = 4k$[/tex], [tex]\( k \)[/tex] must be negative.
- Therefore, [tex]\( k = -\frac{1}{5} \)[/tex].
5. Find the values of [tex]$\sin \theta$[/tex] and [tex]$\cos \theta$[/tex]:
- Substitute [tex]\( k \)[/tex] back into our expressions for [tex]$\sin \theta$[/tex] and [tex]$\cos \theta$[/tex]:
[tex]\[ \sin \theta = -3k = -3 \left(-\frac{1}{5}\right) = \frac{3}{5} = 0.6 \][/tex]
[tex]\[ \cos \theta = 4k = 4 \left(-\frac{1}{5}\right) = -\frac{4}{5} = -0.8 \][/tex]
Therefore, the values of the trigonometric functions are:
[tex]\[ \sin \theta = 0.6 \][/tex]
[tex]\[ \cos \theta = -0.8 \][/tex]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.