Question

Cos210+sin150-tg240=

Answer 1

Cos210+sin150-tg240= (-√3)/2+1/2-√3= 1-√3-√3= 1-2√3

Answer 2

To find the value of $\cos(210^\circ) + \sin(150^\circ) - \tan(240^\circ)$, you can use trigonometric identities and the unit circle.

1. $\cos(210^\circ)$: The cosine function is negative in the third quadrant, and the reference angle for $210^\circ$ is $30^\circ$. So, $\cos(210^\circ) = -\cos(30^\circ) = -\frac{\sqrt{3}}{2}$.

2. $\sin(150^\circ)$: The sine function is positive in the second quadrant, and the reference angle for $150^\circ$ is $30^\circ$. So, $\sin(150^\circ) = \sin(30^\circ) = \frac{1}{2}$.

3. $\tan(240^\circ)$: The tangent function is negative in the third quadrant, and the reference angle for $240^\circ$ is $60^\circ$. So, $\tan(240^\circ) = -\tan(60^\circ) = -\sqrt{3}$.

Now, you can substitute these values into the expression:

$$\cos(210^\circ) + \sin(150^\circ) - \tan(240^\circ) = -\frac{\sqrt{3}}{2} + \frac{1}{2} - (-\sqrt{3}) = -\frac{\sqrt{3}}{2} + \frac{1}{2} + \sqrt{3}$$

To simplify the expression, you can first find a common denominator for the fractions:

$$-\frac{\sqrt{3}}{2} + \frac{1}{2} + \frac{2\sqrt{3}}{2} = -\frac{\sqrt{3} + 2\sqrt{3}}{2} + \frac{1}{2}$$

Combine the fractions:

$$-\frac{3\sqrt{3}}{2} + \frac{1}{2}$$

So, $\cos(210^\circ) + \sin(150^\circ) - \tan(240^\circ) = -\frac{3\sqrt{3}}{2} + \frac{1}{2}$