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Данякеп
@Данякеп
September 2021
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oganesbagoyan
Verified answer
Task/26548397
-------------------.
1.
tgα
=sinα / cosα = -√(1 -cos²α) / cosα =
-√(1 -1/10)
/
1/
√10
= - 3
.
* * * α∈(3π/2 ;2π) , sin
α < 0 * * *
----------------
2. tgα
=sinα / cosα = sinα /( - √(1 -sin²α) ) =( -5/√26 )
/
( -√( 1 -( -5/√26 )²)
= 5.
* * * α∈(π; 3π/2 ) , cos
α < 0 * * *
----------------
3. 3cosα
= 3√(1 -sin²α) ) =3√
(
1 -( -2√2 /3 )²
)
= 3√(1 -8/9) =3√(1/9) =3*1/3
=1.
* * * α∈(3π/2 ; 2π ) , cos
α > 0 * * *
2 votes
Thanks 1
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Answers & Comments
Verified answer
Task/26548397-------------------.
1. tgα =sinα / cosα = -√(1 -cos²α) / cosα = -√(1 -1/10) / 1/√10 = - 3.
* * * α∈(3π/2 ;2π) , sinα < 0 * * *
----------------
2. tgα =sinα / cosα = sinα /( - √(1 -sin²α) ) =( -5/√26 ) / ( -√( 1 -( -5/√26 )²) = 5.
* * * α∈(π; 3π/2 ) , cosα < 0 * * *
----------------
3. 3cosα = 3√(1 -sin²α) ) =3√( 1 -( -2√2 /3 )² ) = 3√(1 -8/9) =3√(1/9) =3*1/3 =1.
* * * α∈(3π/2 ; 2π ) , cosα > 0 * * *