Question
Let α,β be such that π < (α - β) < 3π. If sin α + sin β = -2165 and cos α + cos β = -2765, then the value of \
cos α−β2
cos α−β2
Answer: Option D
:
D
(d)sin α + sin β = -2165 , cosα + cos β = -2765
Now (sin α + sin β)2 + (cosα + cos β)2 = (−2165)2 +(−2765)2
⇒ 2 + 2 sin α sin β + 2 cos α cos β = 441652 + 729652
⇒ 2 + 2[cos (α - β)] = 1170(65)2 ⇒ 2.2 cos2(α+β2)=1170(65)2
⇒cos (α−β2) = 3√130130 = 3√130
Therefore cos (α−β2) = −3√130 , { ∵ π2 < α−β2 < 3π2 }
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:
D
(d)sin α + sin β = -2165 , cosα + cos β = -2765
Now (sin α + sin β)2 + (cosα + cos β)2 = (−2165)2 +(−2765)2
⇒ 2 + 2 sin α sin β + 2 cos α cos β = 441652 + 729652
⇒ 2 + 2[cos (α - β)] = 1170(65)2 ⇒ 2.2 cos2(α+β2)=1170(65)2
⇒cos (α−β2) = 3√130130 = 3√130
Therefore cos (α−β2) = −3√130 , { ∵ π2 < α−β2 < 3π2 }
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