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katykatusha25
@katykatusha25
July 2022
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ПОМОГИТЕ! ТРИГОНОМЕТРИЯ! 1)
1)tg a=15/8, П<a<3П/2 (Найти: cos a, sin a,ctg a) 2)ctg a= - 3, 3П/2<a<2П (Найти: sin a,cos a, tg a)
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oganesbagoyan
Verified answer
1)
tqα=15/8, π<α<3π/2 .
-----------------------------
sinα - ? , cosα-?, ctqα -?
ctqα =
1/tqα =1/(15/8) =
8/15
.
cos α= 1/±√(1+tq²α) .
cos α= 1/(-√(1+tq²α)) = -1/√(1+tq²α) , т.к. , если π<α<3π/2 , то cosα<0 .
cosα =
-1√(1+(15/8)²) = -1√/(1+225/64) = -1/√(289/64)= -1/(17/8) =
- 8/17.
sinα =
cosα*tqα =(-8/17)*(15/8) =
-15/17
.
*********************************
2)
ctqα = -3 ; 3π/2<α< 2π.
----------------------------------
sinα - ? , cosα-?, tqα -?
tqα =
1/ctqα =
-1/3.
sinα = ±1/√(1+ctq²α) .
sinα = -1/√(1+ ctq²α) , т.к. , если π<α<3π/2<α<2π , то sinα<0.
sinα =
-1/√(1+ (-3)²)) = -
1/√10.
cosα=
sinα*ctqα =(-
1/√10)*(-3)=
3/√10
.
************
tqα=sinα/cosα⇒sinα =cosα*tqα.
ctqα =cosα/sinα⇒cosα =sinα*ctqα.
24 votes
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Answers & Comments
Verified answer
1) tqα=15/8, π<α<3π/2 .-----------------------------
sinα - ? , cosα-?, ctqα -?
ctqα = 1/tqα =1/(15/8) =8/15 .
cos α= 1/±√(1+tq²α) .
cos α= 1/(-√(1+tq²α)) = -1/√(1+tq²α) , т.к. , если π<α<3π/2 , то cosα<0 .
cosα = -1√(1+(15/8)²) = -1√/(1+225/64) = -1/√(289/64)= -1/(17/8) = - 8/17.
sinα = cosα*tqα =(-8/17)*(15/8) = -15/17.
*********************************
2) ctqα = -3 ; 3π/2<α< 2π.
----------------------------------
sinα - ? , cosα-?, tqα -?
tqα =1/ctqα = -1/3.
sinα = ±1/√(1+ctq²α) .
sinα = -1/√(1+ ctq²α) , т.к. , если π<α<3π/2<α<2π , то sinα<0.
sinα = -1/√(1+ (-3)²)) = -1/√10.
cosα=sinα*ctqα =(-1/√10)*(-3)=3/√10.
************
tqα=sinα/cosα⇒sinα =cosα*tqα.
ctqα =cosα/sinα⇒cosα =sinα*ctqα.