1) tqα=15/8, π<α<3π/2 .<br>-----------------------------
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 . <br>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.
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2) ctqα = -3 ; 3π/2<α< 2π.<br>----------------------------------
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.<br>sinα = -1/√(1+ (-3)²)) = -1/√10.
cosα=sinα*ctqα =(-1/√10)*(-3)=3/√10.
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tqα=sinα/cosα⇒sinα =cosα*tqα.
ctqα =cosα/sinα⇒cosα =sinα*ctqα.