The proof of inequality (sinA)^2+(sinB)^2=5(sinC)^2 A. B and C are the three inner angles of a triangle Verification: sinc is less than or equal to 0.6

After both sides of the original equation are multiplied by R ^ 2 (the square of the radius of the circumscribed circle of the triangle), we can get it from the sine theorem
a^2+b^2=5c^2,.①
Where a, B and C represent the opposite sides of angles a, B and C respectively
From the cosine theorem:
a^2+b^2-2abcosC=c^2.②
① If we subtract formula 2, we get 2C ^ 2 = abcosc
==0.8
And (sinc) ^ 2 + (COSC) ^ 2 = 1
So sinc=

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