Dado sen a=1/2 e sen b=1/4.com 0<a,b<Pi/2,determine cos(2a+2b).
Valendo 5* alguem ajuda ae?
cos (2a + 2b) =
= 1 - 2*sen²a + 1 - 2*sen²b
= 1 - 2*1/4 + 1 - 2*1/8
= 1 - 2/4 + 1 - 2/8
= 4 - 2 / 4 + 8 - 2 / 8
= 2/4 + 6/8
= 8/8 = 1
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cos (2a + 2b) =
= 1 - 2*sen²a + 1 - 2*sen²b
= 1 - 2*1/4 + 1 - 2*1/8
= 1 - 2/4 + 1 - 2/8
= 4 - 2 / 4 + 8 - 2 / 8
= 2/4 + 6/8
= 8/8 = 1