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If A = 45^{0 }, verify that:

(ii) cos 2A = 2 cos^{2} A – 1 = 1 – 2 sin^{2} A

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#### Solution

A= 45^{0 }

^{`⇒ 2A = 2xx45^0=90^0`}

(ii) cos 2 A = cos `90^0 = 0`

`2 cos^2 -1 = 2 cos ^2 45 ^0-1 = 2 xx(1/sqrt(2))^2 -1=2 xx1/2 -=1-1=0`

Now , `1-2 sin^2 A =1-2 xx(1/sqrt(2)^2 )-1=1-2xx1/2=1-1=0`

∴ cos 2A = 2 cos^{2} A – 1 = 1 – 2 sin^{2} A

Concept: Trigonometric Ratios and Its Reciprocal

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