हिंदी

If A = 45°, verify that cos 2A = 2 cos^2A – 1 = 1 – 2 sin^2A.

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प्रश्न

If A = 45°, verify that cos 2A = 2 cos2A – 1 = 1 – 2 sin2A.

योग
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उत्तर

A= 45

`⇒ 2A = 2xx45^0=90^0`

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 cos2 A – 1 = 1 – 2 sin2

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अध्याय 11: T-Ratios of Some Particular Angles - EXERCISE 11 [पृष्ठ ५७२]

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आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 11 T-Ratios of Some Particular Angles
EXERCISE 11 | Q 12. (ii) | पृष्ठ ५७२
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