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Given:
[tex]\[
\cos(A) = -\frac{7}{9} \quad \text{and} \quad \sin(A) \ \textless \ 0.
\][/tex]

Find \(\sin(A)\) and \(\tan(A)\).

A. \(\sin(A) = -4\sqrt{2}, \tan(A) = -\frac{4\sqrt{2}}{7}\)

B. \(\sin(A) = -\frac{4\sqrt{2}}{9}, \tan(A) = \frac{4\sqrt{2}}{7}\)

C. \(\sin(A) = 4\sqrt{2}, \tan(A) = \frac{4\sqrt{2}}{9}\)

D. [tex]\(\sin(A) = -\frac{4\sqrt{2}}{9}, \tan(A) = \frac{4\sqrt{2}}{9}\)[/tex]


Sagot :

To solve this trigonometric problem, follow these steps:

1. Understand the Quadrant:
Given \(\cos(A) = -\frac{7}{9}\) and \(\sin(A) < 0\), angle \(A\) is in the third quadrant. In the third quadrant, both \(\cos\) and \(\sin\) are negative, and \(\tan\) is positive.

2. Find \(\sin(A)\):
We use the Pythagorean identity for sine and cosine:
[tex]\[ \sin^2(A) + \cos^2(A) = 1 \][/tex]
Substitute \(\cos(A) = -\frac{7}{9}\) into the equation:
[tex]\[ \sin^2(A) + \left(-\frac{7}{9}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2(A) + \left(\frac{49}{81}\right) = 1 \][/tex]
Solve for \(\sin^2(A)\):
[tex]\[ \sin^2(A) = 1 - \frac{49}{81} \][/tex]
[tex]\[ \sin^2(A) = \frac{81}{81} - \frac{49}{81} \][/tex]
[tex]\[ \sin^2(A) = \frac{32}{81} \][/tex]
Take the square root and select the negative value (since \(\sin(A)\) is negative in the third quadrant):
[tex]\[ \sin(A) = -\sqrt{\frac{32}{81}} \][/tex]
Simplify the square root:
[tex]\[ \sin(A) = -\frac{\sqrt{32}}{9} \][/tex]
[tex]\[ \sin(A) = -\frac{4\sqrt{2}}{9} \][/tex]

3. Find \(\tan(A)\):
The tangent function is the ratio of the sine and cosine functions:
[tex]\[ \tan(A) = \frac{\sin(A)}{\cos(A)} \][/tex]
Substitute \(\sin(A) = -\frac{4\sqrt{2}}{9}\) and \(\cos(A) = -\frac{7}{9}\):
[tex]\[ \tan(A) = \frac{-\frac{4\sqrt{2}}{9}}{-\frac{7}{9}} \][/tex]
Simplify the expression:
[tex]\[ \tan(A) = \frac{4\sqrt{2}}{7} \][/tex]

Now let's match the computed values with the given options:
- \(\sin(A) = -\frac{4\sqrt{2}}{9}\)
- \(\tan(A) = \frac{4\sqrt{2}}{7}\)

The answer matches the second choice:

[tex]\[ \sin (A)=-\frac{4 \sqrt{2}}{9}, \tan (A)=\frac{4 \sqrt{2}}{7} \][/tex]
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