Question

12. If A= 45, verify that:
(i) sin 2A = 2 sin A cos A
(u) cos 2A =2cos A-1=1-2 sin A​

Answer 1

Answer:

i)

RHS:

2*sin45*cos45 = 2*(1/root2)*(1/root2) = 2*1/2 = 1

LHS:

sin(2*45) = sin90 = 1

Hence LHS = RHS

ii)

LHS:

cos(2*45) = cos90 = 0

MIDDLE TERM:

2cos^2(45)-1 = 2*(1/root2)squared - 1 = 2*(1/2) - 1 = 0

RHS:

1 - 2sin^2(45) = 1 - 2*(1/root2)squared = 1 - 2*(1/2) = 0

Hence LHS = MIDDLE TERM = RHS.

Please mark me the brainliest ( I am writing this at 2:53 AM)

Answer 2

Step-by-step explanation:

$a = 45 \\ \sin(45 ) = 1 \div \sqrt{2} \\ \cos(45) = 1 \div \sqrt{2} \\ i)sin2a = \sin(90) = 1 = lhs \\ rhs = 2 \sin(45) \times \cos(45) = 2 \times( 1 \div \sqrt{2} ) \times (1 \div \sqrt{2} ) \\ = 2 \div 2 = 1 = rhs \\ hence \: lhs = rhs \\ \\ ii) \cos(2a) = 2 \cos(a) - 1 = 1 - 2 \sin(a) \\ \cos(2a) = \cos(90) = 0 \\ 2 \cos(a) - 1 = 2 \div \sqrt{2} - 1 \: ie \: not \: equals \: to \: zeohne \: this can \: notbe \: proved$

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