Question

здесь несколько заданий:1) tg 5p/42) cos (5/4)p3)cos (5/3)p4)sin(-13/3)pпожалуйста срочно надо=(

Answer 1

1) $tg(\frac{5\pi}{4})=tg(\pi+\frac{\pi}{4})=tg(\frac{\pi}{4})=1$

2) $cos(\frac{5\pi}{4})=cos(\pi+\frac{\pi}{4})=-cos(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$

3) $cos(\frac{5\pi}{3})=cos(2\pi-\frac{\pi}{3})=-cos(\frac{\pi}{3})=-\frac{1}{2}$

4) $sin(\frac{-13\pi}{3})=- sin(\frac{13\pi}{3}) =-sin(4\pi+\frac{\pi}{3})=-sin(\frac{\pi}{3})=-\frac{\sqrt{3}}{2}$  

Answer 2

1) To solve the equation tg(5π/4), we can use the trigonometric identity tan(x) = sin(x)/cos(x). So, we have:

tg(5π/4) = sin(5π/4) / cos(5π/4)

Using the unit circle, we can determine the values of sin(5π/4) and cos(5π/4). At 5π/4, the terminal point is located in the third quadrant, where both sin and cos are negative. Therefore:

sin(5π/4) = -√2/2 cos(5π/4) = -√2/2

Substituting these values into the equation, we get:

tg(5π/4) = (-√2/2) / (-√2/2)

The negative signs cancel out, and we are left with:

tg(5π/4) = √2/2

So, the solution to tg(5π/4) is √2/2.

2) To solve the equation cos(5π/3), we can again use the unit circle. At 5π/3, the terminal point is located in the second quadrant, where cos is negative. Therefore:

cos(5π/3) = -1/2

So, the solution to cos(5π/3) is -1/2.

3) To solve the equation cos(5π/3), we can use the unit circle. At 5π/3, the terminal point is located in the second quadrant, where cos is negative. Therefore:

cos(5π/3) = -1/2

So, the solution to cos(5π/3) is -1/2.

4) To solve the equation sin(-13π/3), we can use the unit circle. At -13π/3, the terminal point is located in the fourth quadrant, where sin is negative. Therefore:

sin(-13π/3) = -√3/2

So, the solution to sin(-13π/3) is -√3/2.

I hope this helps! Let me know if you have any further questions.