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If
4
cos
36
∘
+
cot
(
7
1
2
∘
)
=
n
1
+
n
2
+
n
3
+
n
4
+
n
5
+
n
6
4 \cos 36^\circ + \cot \left( 7 \frac{1}{2}^\circ \right) = \sqrt{n_1} + \sqrt{n_2} + \sqrt{n_3} + \sqrt{n_4} + \sqrt{n_5} + \sqrt{n_6}
4
cos
3
6
∘
+
cot
(
7
2
1
∘
)
=
n
1
+
n
2
+
n
3
+
n
4
+
n
5
+
n
6
…
Mathematics
Problem
If
4
cos
36
∘
+
cot
(
7
1
2
∘
)
=
n
1
+
n
2
+
n
3
+
n
4
+
n
5
+
n
6
4 \cos 36^\circ + \cot \left( 7 \frac{1}{2}^\circ \right) = \sqrt{n_1} + \sqrt{n_2} + \sqrt{n_3} + \sqrt{n_4} + \sqrt{n_5} + \sqrt{n_6}
4
cos
3
6
∘
+
cot
(
7
2
1
∘
)
=
n
1
+
n
2
+
n
3
+
n
4
+
n
5
+
n
6
, then the value of
∑
i
=
1
6
n
i
2
\sum_{i=1}^{6} n_i^2
∑
i
=
1
6
n
i
2
must be
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