We are given that $\cos(2\alpha) = -4781$ (which seems very unusual because the cosine of an angle cannot exceed 1 or be less than -1) and that $\frac{\pi}{2} < 2\alpha < \pi$. Let’s assume this was a typo, and the more reasonable scenario might be that $\cos(2\alpha) = -\frac{4}{5}$, which is within the valid range for cosine and provides a practical problem. Let’s proceed with this assumption and solve it:

Step 1: Use $\cos(2\alpha) = 2\cos^2(\alpha) - 1$ to find $\cos(\alpha)$

We know the double angle identity for cosine:

$\cos(2\alpha) = 2\cos^2(\alpha) - 1$

Substitute $\cos(2\alpha) = -\frac{4}{5}$ into the equation:

$-\frac{4}{5} = 2\cos^2(\alpha) - 1$

Solve for $\cos^2(\alpha)$:

$2\cos^2(\alpha) = -\frac{4}{5} + 1 = \frac{1}{5}$

$\cos^2(\alpha) = \frac{1}{10}$

Thus,

$\cos(\alpha) = \pm \frac{1}{\sqrt{10}}$

Step 2: Determine the sign of $\cos(\alpha)$

Since $\frac{\pi}{2} < 2\alpha < \pi$, the angle $\alpha$ lies in the second quadrant. In the second quadrant, cosine is negative, so:

$\cos(\alpha) = -\frac{1}{\sqrt{10}}$

Step 3: Find $\sin(\alpha)$

Use the Pythagorean identity $\sin^2(\alpha) + \cos^2(\alpha) = 1$ to find $\sin(\alpha)$:

$\sin^2(\alpha) = 1 - \cos^2(\alpha) = 1 - \frac{1}{10} = \frac{9}{10}$

Thus,

$\sin(\alpha) = \pm \frac{3}{\sqrt{10}}$

Since $\alpha$ is in the second quadrant, sine is positive, so:

$\sin(\alpha) = \frac{3}{\sqrt{10}}$

Step 4: Find the remaining trigonometric functions

• Tangent: $\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{3}{\sqrt{10}}}{-\frac{1}{\sqrt{10}}} = -3$

• Cosecant: $\csc(\alpha) = \frac{1}{\sin(\alpha)} = \frac{\sqrt{10}}{3}$

• Secant: $\sec(\alpha) = \frac{1}{\cos(\alpha)} = -\sqrt{10}$

• Cotangent: $\cot(\alpha) = \frac{1}{\tan(\alpha)} = -\frac{1}{3}$

Final Answer:

• $\sin(\alpha) = \frac{3}{\sqrt{10}}$
• $\cos(\alpha) = -\frac{1}{\sqrt{10}}$
• $\tan(\alpha) = -3$
• $\csc(\alpha) = \frac{\sqrt{10}}{3}$
• $\sec(\alpha) = -\sqrt{10}$
• $\cot(\alpha) = -\frac{1}{3}$

Would you like more details or have any questions?

Follow-up questions:

1. How would the solution change if $\alpha$ were in the first quadrant?
2. What is the geometric interpretation of the given trigonometric values?
3. Can you solve for $\alpha$ exactly based on these values?
4. How would you find the values of the trigonometric functions for $2\alpha$ directly?
5. What would happen if $\cos(2\alpha)$ had a different sign?

Tip:

Always remember which quadrant your angle is in, as it determines the signs of sine, cosine, and tangent.