The image contains several trigonometric problems. I will analyze and solve each problem step by step. Let's proceed with the solutions:

5. $\sin 2\alpha$, если $\cos \alpha = 0.6$, $\pi < \alpha < 2\pi$:

• $\cos \alpha = 0.6$
  Using $\sin^2 \alpha + \cos^2 \alpha = 1$:
  $\sin^2 \alpha = 1 - (0.6)^2 = 1 - 0.36 = 0.64$, so $\sin \alpha = \pm 0.8$.
  Since $\pi < \alpha < 2\pi$, $\sin \alpha$ is negative: $\sin \alpha = -0.8$.

• $\sin 2\alpha = 2 \sin \alpha \cos \alpha = 2(-0.8)(0.6) = -0.96$.

Answer: $\sin 2\alpha = -0.96$.

6. $\frac{37}{\sin^2 173^\circ + \sin^2 263^\circ}$:

• $\sin^2 173^\circ = \sin^2 (180^\circ - 7^\circ) = \sin^2 7^\circ$.
• $\sin^2 263^\circ = \sin^2 (360^\circ - 97^\circ) = \sin^2 (-97^\circ) = \sin^2 97^\circ$.
• Using $\sin^2 x + \sin^2 (90^\circ + x) = 1$:
  $\sin^2 7^\circ + \sin^2 97^\circ = 1$.

Answer: $\frac{37}{1} = 37$.

7. $20\sqrt{3} \cos \left(-\frac{\pi}{3}\right) \sin\left(-\frac{\pi}{3}\right)$:

• $\cos\left(-\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$.
• $\sin\left(-\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}$.
• Substituting:
  $20\sqrt{3} \cdot \frac{1}{2} \cdot -\frac{\sqrt{3}}{2} = 20\sqrt{3} \cdot -\frac{3}{4} = -15\sqrt{3}$.

Answer: $-15\sqrt{3}$.

8. $59 \tan 56^\circ \cdot \tan 34^\circ$:

• Using $\tan x \tan (90^\circ - x) = 1$:
  $\tan 56^\circ \cdot \tan 34^\circ = 1$.
• Substituting: $59 \cdot 1 = 59$.

Answer: $59$.

9. $\frac{13 \sin 152^\circ}{\cos 76^\circ \cdot \cos 14^\circ}$:

• $\sin 152^\circ = \sin (180^\circ - 28^\circ) = \sin 28^\circ$.
• $\cos 76^\circ = \sin 14^\circ$ (since $\cos x = \sin (90^\circ - x)$).
• Substituting:
  $\frac{13 \sin 28^\circ}{\sin 14^\circ \cdot \cos 14^\circ}$.
• Using $\sin 2x = 2 \sin x \cos x$:
  $\sin 14^\circ \cdot \cos 14^\circ = \frac{\sin 28^\circ}{2}$.
  So, $\frac{13 \sin 28^\circ}{\frac{\sin 28^\circ}{2}} = 13 \cdot 2 = 26$.

Answer: $26$.

Let me know if you'd like me to proceed with the remaining problems or explain any solution in more detail.

5 Related Questions:

1. How can we determine the sign of trigonometric functions in different quadrants?
2. Why does $\sin^2 x + \cos^2 x = 1$ hold for all values of $x$?
3. What is the significance of complementary angles in trigonometry?
4. How is $\sin(180^\circ - x)$ related to $\sin x$?
5. How can we simplify expressions involving double angles?

One Tip:

Always verify the quadrant of the angle to correctly assign the signs for sine, cosine, or tangent.