Question

3sin30°-2ctg45°+cos60°+5tg180°

Answer 1

$3sin(30^{\circ})-2ctg(45^{\circ})+cos(60^{\circ})+5tg(180^{\circ})\\3\cdot \frac{1}{2}-2\cdot 1+\frac{1}{2}+5\cdot 0=\frac{3}{2}+\frac{1}{2}-2=0$

Answer 2

Let's evaluate the expression step by step:

1. First, calculate the values of the trigonometric functions for the given angles:

• sin(30°) = 0.5
• ctg(45°) is the cotangent of 45°, which is equal to 1.
• cos(60°) = 0.5
• tan(180°) is the tangent of 180°. The tangent function has a period of 180°, so tan(180°) is the same as tan(0°), which is 0.

2. Now, substitute these values into the expression:

3sin(30°) - 2ctg(45°) + cos(60°) + 5tg(180°) = 3(0.5) - 2(1) + 0.5 + 5(0)

3. Perform the calculations:

1.5 - 2 + 0.5 + 0 = 0.

So, the value of the expression is 0.