Question

6sin(-90°)-3cos(-270°)+4ctg(-45°)​

Answer 1

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Answer 2

Let's evaluate each term one by one:

1. $6\sin(-90^\circ)$

The sine function has a period of 360 degrees, and $\sin(-90^\circ)$ is equal to $-1$ because the sine function is negative in the fourth quadrant. So,

$6\sin(-90^\circ) = 6(-1) = -6$.

2. $3\cos(-270^\circ)$

The cosine function also has a period of 360 degrees, and $\cos(-270^\circ)$ is equal to $-1$ because the cosine function is negative in the third quadrant. So,

$3\cos(-270^\circ) = 3(-1) = -3$.

3. $4\cot(-45^\circ)$

The cotangent function is the reciprocal of the tangent function, and $\cot(-45^\circ)$ can be calculated as the reciprocal of $\tan(-45^\circ)$. The tangent of $-45^\circ$ is $-1$, so

$\cot(-45^\circ) = \frac{1}{\tan(-45^\circ)} = \frac{1}{-1} = -1$.

Now, let's put it all together:

$6\sin(-90^\circ) - 3\cos(-270^\circ) + 4\cot(-45^\circ) = -6 - (-3) + (-4) = -6 + 3 - 4 = -7$.

So, the final result is $-7$.