IPMAT Indore 2022 - QA (SA) Q4 Explanation

Explanation:

100)

Given sinα  + sinβ = √2/√3

Squaring both sides, we get (sinα  + sinβ)^2 + sin^2β + 2sinα.sinβ = 2/3 ...........(1)

&

cosα + cosβ  = 1/√3

(cosα + cosβ)^2 = (1/√3)^2

= cos^2α + cos^2β + 2cosα.cosβ = 1/3 ..............(2)

Adding equations (1) and (2), we get

(sinα  + sinβ)^2 + sin^2β + 2sinαsinβ + cos^2α + cos^2β + 2cosα.cosβ 

= 1

= 1 + 1 + 2 [sinα.sinβ + cosα.cosβ] = 1

= 2 cos(α - β) = -1

= cos(α - β) = −1/2

= 2 cos^2 (α−β/2) -1 = −1/2 [∵ cos2A = cos2A-1]

= 2 cos^2 (α−β/2) = 1/2

= cos^2 (α−β/2) = 1/4

We have to find the values of (20cos(α−β/2))^2

(20cos(α−β/2))^2 = (20)^2 cos^2(α−β/2)

= 400 × 1/4

= 100.