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Question : 21 of 160
Marks:
+1,
-0
Solution:
Given,
α+β=γ Then,
cos2α+cos2β+cos2γ =[2cos2α+2cos2β+2cos2γ] =[1+cos‌2‌α+1+cos‌2‌β+2cos2γ] [∵2cos2A=1+cos‌2‌A] =[2+cos‌2‌α+cos‌2‌β+2cos2γ] =[2+2‌cos()‌cos(α−β)+2cos2γ] =[2+2‌cos‌γ‌cos(α−β)+2cos2γ] =1+cos‌γ‌cos(α−β)+cos2γ =1+cos‌γ{cos(α−β)+cos‌γ} [Put
γ=α+β]
=1+2‌cos‌γ‌cos().cos(−) =1+2‌cos‌γ‌cos‌α.cos(−β) =1+2‌cos‌α‌cos‌β‌cos‌γ [∵cos(−θ)=cos‌θ]
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