Вычислите
a) sinα, если cos(α+ π/3)= - 3/5 , π/2 < α+π/3 < π .<br>---
Т.к. π/2 < α+π/3 < π , то <strong>sin(α+ π/3) = √(1-cos²(α+ π/3) =√(1 -(-3/5)² ) = 4/5.
cos(α+ π/3)= - 3/5 ⇔cosα*cosπ/3 - sinα*sinπ/3 = - 3/5⇔
(1/2)*cosα - (√3/2)*sinα = - 3/5 ⇔ -(√3/2)*cosα + (3/2)*sinα = (3√3)/5 (1)
sin(α+ π/3) = 4/5 ⇔sinα*cosπ/3 +cosα*sinπ/3=4/5 ⇔
(√3/2)*cosα +(1/2)*sinα= 4/5 (2)
складывая (1) и (2) получаем:
sinα = (4 +3√3) /10.
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b) cosα, если sin(π/6-α)= 2√2/3, π/2<π/6 - α < <span>π .
Т.к. π/2 < π/6 - α < π , то <strong>cos(π/6-α) = - √(1- sin² (π/6-α) )= -1/3.
sin(π/6-α)= 2√2/3⇔sinπ/6*cosα -cosπ/6*sinα =2√2/3⇔
(1/2)*cosα - √3/ 2*sinα = 2√2/3 (1) .
cos(π/6-α) = -1/3 ⇔ cosπ/6*cosα+sinπ/6*sinα= -1/3⇔
√3/2*cosα +1/2*sinα = -1/3 ⇔ 3/2*cosα +√3/2*sinα = - √3 /3 (2).
складывая (1) и (2) получаем:
cosα = (2√2 - √3)/ 6 .