The value of $1 + cos 56^{o} + cos 58^{o} - cos 66^{o}$ is equal to

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Solution

$1 + c o s 56 = 2 c o s 228 a s c o s 2 t$

$= 2 c o s 2 t - 1 c o s 58$

$= 2 c o s 29 - 1 c o s 66$

$= 2 c o s 233 - 1$
so
$= 1 + c o s 56 + c o s 58 - c o s 66$

$= 2 (c o s 228 + c o s 229 - c o s 233)$

Now

$28 + 29 + 33 = 90$

In case we demonstrate

$c o s 2 A + c o s 2 B - c o s 2 c$

$= 2 c o s A c o s B s i n C$

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