This is the basic and conceptual question from trigonometric ratios in which we have given A = 45°
We have to prove that cos 2A = 2cos^2 A -1 = 1 – sin^2 A
RS Aggarwal, Class 10, chapter 11, question no 12(ii)
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A=45∘ then 2A=90∘
cos2A = cos90∘ =0
2cos^2A−1 = 2cos^2 45∘−1
=2(1/√2)^2−1
=1−1
=0
1−sin^2A
=1−2sin^2 45∘
=1−2×(1/√2)^2
=1−1
=0
∴cos 2A =2cos^2A−1 =1−2sin^2A.