Sin(arcctg (1/2) - arcctg(-1/3))^2

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Let's first recall the trigonometric identity that relates the tangent and cotangent functions:

[\text{arccot}(x) = \arctan\left(\frac{1}{x}\right)]

So, we can rewrite the given expression as:

[\sin\left(\arctan(2) - \arctan\left(-\frac{1}{3}\right)\right)^2]

Now, let's use the trigonometric identity for the difference of angles:

[\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta]

Applying this identity, we get:

[\sin(\arctan(2))\cos(\arctan(-1/3)) - \cos(\arctan(2))\sin(\arctan(-1/3))]

Next, we use the trigonometric identities:

[\sin(\arctan(x)) = \frac{x}{\sqrt{1+x^2}} \hspace{0.5cm}\text{and}\hspace{0.5cm} \cos(\arctan(x)) = \frac{1}{\sqrt{1+x^2}}]

Substitute these into the equation above:

[\frac{2}{\sqrt{5}}\cdot\frac{3}{\sqrt{10}} - \frac{1}{\sqrt{5}}\cdot \frac{-1}{\sqrt{10}}]

We simplify the expression:

[\frac{6}{\sqrt{50}} + \frac{1}{\sqrt{50}} = \frac{7}{\sqrt{50}} = \frac{7\sqrt{50}}{50}]

Therefore, the value of (\sin(\text{arccot}(1/2) - \text{arccot}(-1/3))^2) is (\frac{7\sqrt{50}}{50}).