jika α + β + γ = 180°, buktikan bahwa sin^2α + sin^2β + sin^2β = 2(1+cosα cosβ cosγ) mohon bantuannya ka

cat :
>> 2 cosx . cosy = cos(x+y) + cos(x-y)
>> sin²x = 1/2 (1 - cos2x)
>> cos(180 - x) = -cosx
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jika a + b + c = 180
maka
sin²a + sin²b + sin²c
= 1/2 (1 - cos2a) + 1/2 (1 - cos2b) + 1/2 (1 - cos2c)
= 1/2 (3 - [cos2a + cos2b + cos2c] )
= 1/2 (3 - [2 cos(a+b).cos(a-b) + cos2c] )

c = 180 - (a+b)
cos(c) = cos(180 - (a+b))
           = -cos(a+b)

= 1/2 (3 - [- 2 cos(c).cos(a-b) + cos2c] )

cos(2c) = 2cos²c - 1

= 1/2 (3 - [- 2 cos(c).cos(a-b) + 2cos²c - 1] )
= 1/2 (3 + 2 cos(c).cos(a-b) - 2cos²c + 1)
= 1/2 (4 + 2 cos(c).cos(a-b) - 2cos²c)
= 2 +  cos(c).cos(a-b) - cos²c
= 2 + cos(c) [cos(a-b) - cos(c)]
= 2 + cos(c) [cos(a-b) + cos(a+b)]
= 2 + cos(c) [2 cos(a) . cos(b)]
= 2 (1 + cos(a) . cos(b) . cos(c) )   [terbukti]