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daniilgalaev
@daniilgalaev
June 2021
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Друзья, помогите, пожалуйста решить пример:
-cos(2arctg4/3)
Заранее СПАСИБО!!
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oganesbagoyan
Verified answer
-π/2 < arctq4/3 < π/2 ; * * * - π/2 < arctqa < π/2 * * *
α =arctq4/3⇒tqα =tq(arctq4/3) = 4/3 .
cosα =1/√(1+tq²α) =1/√(1+(4/3)²) =3/5.
------------
- cos(2α) = -(2cos²α - 1) =1 -2cos²α =1 - 2(3/5)²
=7/25
.
2 votes
Thanks 1
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Answers & Comments
Verified answer
-π/2 < arctq4/3 < π/2 ; * * * - π/2 < arctqa < π/2 * * *α =arctq4/3⇒tqα =tq(arctq4/3) = 4/3 .
cosα =1/√(1+tq²α) =1/√(1+(4/3)²) =3/5.
------------
- cos(2α) = -(2cos²α - 1) =1 -2cos²α =1 - 2(3/5)² =7/25.