Question

Cos135°+tg225°-sin(-210°)

Answer 1

Ответ:

$\cos(135°) + \tan(225°) - \sin( - 210°) = - \frac{ \sqrt{2} }{2} + 1 - \frac{1}{2} = - \frac{ \sqrt{2} }{2} + \frac{1}{2} = \frac{ - \sqrt{2 + 1} }{2}$

Answer 2

Cos135°: To find the cosine of 135 degrees, we can use the fact that the cosine function is periodic with a period of 360 degrees. We can find the reference angle by subtracting 90 degrees from 135, which gives us 45 degrees. Since 135 degrees is in the second quadrant, the cosine of 135 degrees is equal to the negative of the cosine of its reference angle. Therefore, cos(135°) = -cos(45°) = -√2/2.

Tg225°: To find the tangent of 225 degrees, we can use the fact that the tangent function is periodic with a period of 180 degrees. We can find the reference angle by subtracting 180 degrees from 225, which gives us 45 degrees. Since 225 degrees is in the third quadrant, the tangent of 225 degrees is equal to the negative of the tangent of its reference angle. Therefore, tg(225°) = -tg(45°) = -1.

Sin(-210°): To find the sine of -210 degrees, we can use the fact that the sine function is an odd function, meaning that sin(-x) = -sin(x). Therefore, sin(-210°) = -sin(210°). Since the sine function is periodic with a period of 360 degrees, sin(210°) is equal to sin(210° - 360°) = sin(-150°). Since -150 degrees is in the fourth quadrant, the sine of -150 degrees is positive. Therefore, sin(-210°) = sin(-150°) = √3/2.

Putting it all together: Cos135° + tg225° - sin(-210°) = (-√2/2) + (-1) - (√3/2) = -√2/2 - 1 - √3/2 ≈ -3.366.