Question

"Question24
If A = 45° , verify that : cos 2 A = 2 cos²A - 1 = 1 - 2 sin²A
Chapter6,T-Ratios of particular angles Exercise -6 ,Page number 289"

Answer 1

Hey there!

Given to verify : cos 2 A = 2 cos²A - 1 = 1 - 2 sin²A

If A = 45°

1) cos 2A = cos (2*45) = cos90 = 0 [ T - Ratio of 90° , π/2 ]
2) sinA = sin45 = 1/√2 [ T - Ratio of 45° , π/4 ]
3)cosA = cos45 = 1/√2 [ T - Ratio of 45° , π/4 ]

Now,
Given equation to verify : cos 2 A = 2 cos²A - 1 = 1 - 2 sin²A

From this, cos2A = 2cos²A - 1 , cos2A = 1 -2 sin²A.

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Finding the value of cos2A,

= cos ( 2A )

= cos90

= 0

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Finding the value of 2cos²A - 1

= 2 cos²45 - 1
= 2 ( 1/√2)² - 1
= 2 ( 1/2) - 1
= 1 - 1
= 0

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Finding the value of 1 - 2sin²A
= 1 - 2sin²45
= 1 - 2 ( 1/√2)²
= 1 - 2/2
= 1 - 1
= 0

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Here, We observe that cos2A = 1 - 2sin²A = 2cos²A - 1

Hence proved that, cos 2 A = 2 cos²A - 1 = 1 - 2 sin²A for A = 45°

Answer 2

HELLO DEAR,


GIVEN THAT:-

A = 45°


NOW,

GIVEN Equation is,

cos2A = 2cos²A - 1 = 1 - 2sin²A

NOW substitute the values of "A"

we get,

cos(2 * 45°) = cos90°

cos90° = 0-------(1),

2cos²A - 1 = 2cos²45° - 1

= 2 * (1/√2)² - 1

= 2 * 1/2 - 1 = 0------(2)



1 - 2sin²A

= 1 - 2*sin²45°

= 1 - 2*(1/√2)²

= 1 - 1 = 0-----(3)



from---(1) , ---(2) & ---(3)


we get,

cos 2 A = 2 cos²A - 1 = 1 - 2 sin²A



I HOPE ITS HELP YOU DEAR,
THANKS