Copilot 에게도 물어 봤습니다.
True! The given equation holds. Both sides are equal. Here’s the breakdown:
1. Start with the left-hand side (LHS):
-π+arctan1153 -\pi + \arctan\left(\frac{11}{5\sqrt{3}}\right)
2. Simplify the fraction:
1153=11315 \frac{11}{5\sqrt{3}} = \frac{11\sqrt{3}}{15}
3. Use the identity:
arctan(x)+arctan(1/x)=π2 \arctan(x) + \arctan(1/x) = \frac{\pi}{2}
Applying this to our expression:
arctan11315+arctan15113=π2 \arctan\left(\frac{11\sqrt{3}}{15}\right) + \arctan\left(\frac{15}{11\sqrt{3}}\right) = \frac{\pi}{2}
4. Simplify the second term:
15113=5311 \frac{15}{11\sqrt{3}} = \frac{5\sqrt{3}}{11}
5. Combine the arctan terms:
arctan11315+arctan5311=π2 \arctan\left(\frac{11\sqrt{3}}{15}\right) + \arctan\left(\frac{5\sqrt{3}}{11}\right) = \frac{\pi}{2}
6. Use the identity again:
arctan(x)+arctan(y)=arctanx+y1-xy \arctan(x) + \arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)
arctan113+531-(113)(53)=π2 \arctan\left(\frac{11\sqrt{3} + 5\sqrt{3}}{1 - (11\sqrt{3})(5\sqrt{3})}\right) = \frac{\pi}{2}
7. Simplify the numerator:
113+53=163 11\sqrt{3} + 5\sqrt{3} = 16\sqrt{3}
And the denominator:
1-(113)(53)=1-165=-164 1 - (11\sqrt{3})(5\sqrt{3}) = 1 - 165 = -164
8. Final result:
arctan163-164=π2 \arctan\left(\frac{16\sqrt{3}}{-164}\right) = \frac{\pi}{2}
Since 163-164=3-41 \frac{16\sqrt{3}}{-164} = \frac{\sqrt{3}}{-41}
, we have:
arctan3-41=π2 \arctan\left(\frac{\sqrt{3}}{-41}\right) = \frac{\pi}{2} Therefore, the LHS equals -π/2 -\pi/2
.
9. Now let’s look at the right-hand side (RHS):
-π/2-arctan5311 -\pi/2 - \arctan\left(\frac{5\sqrt{3}} {11}\right)
10. We already found that arctan5311=π2 \arctan\left(\frac{5\sqrt{3}}{11}\right) = \frac{\pi}{2}
So the RHS also equals -π/2 -\pi/2
Since both sides are equal, the given equation is true! �
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Copilot 에게도 물어 봤습니다.
True! The given equation holds. Both sides are equal. Here’s the breakdown:
1. Start with the left-hand side (LHS):
2. Simplify the fraction:
3. Use the identity:
Applying this to our expression:
4. Simplify the second term:
5. Combine the arctan terms:
6. Use the identity again:
Applying this to our expression:
7. Simplify the numerator:
And the denominator:
8. Final result:
Since
, we have:
Therefore, the LHS equals
.
9. Now let’s look at the right-hand side (RHS):
10. We already found that
So the RHS also equals
.
Since both sides are equal, the given equation is true! �