#Point out my error pls. (Conditional Trigonometric Identities)
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Yes.
If: A+B+C=π
Prove that Sin²2A+sin²2B+sin²2C = ...
The R.H.S is incorrect in the question
It fails Value in LHS and Value in RHS when I substitute values of A, B, C ; a+b+c=π
hmm
i will check it out
ok
A,B,C are angles of a triangle
there is no solid geometric proof however
@vale forum
@dense orchid
It is meant to be solved the same way as other Conditional Trigonometric Identities
But this one doesn't satisfy RHS and LHS when I substitute random values that sum to 180°

It gives me the same result but when I substitute values such that A+B+C=180°
LHS≠RHS
??
cos(2A+2B) should equal cos(2C) if A+B+C is 180?
check the pair (90,45,45)
2A+2B+2C=2π
2A+2B = 2π-2C
cos(2A+2B)=cos(2π-2c)=> ....
which equals cos(2C)
I misunderstand 2π as 180 😭😭😭😭
cos is an even function
oh
star
I forgot :(
Cos is +ve in 4th quad.
your solution is correct
but my first step is incorrect :(
It works for certain values
Like 0,0,180 or something
but not all
sin²(2A)+sin²(2B)+sin²(2C)=2[1-cos(2A)cos(2B)cos(2C)] ; [A+B+C=π rad]
Finally solved ✅