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Proof of trig double angle formulae sin ( 2A ) & cos ( 2A )

B.Sudhakar
31/12/2016 0 1

sin (A + B)= sin A cos B + cos A sin B

Put B=A in the above formula 

= Sin ( A + A ) = sin A cos A + cos A sin A

= 2 sin A cos A

∴ sin ( 2 A )= 2 sin A cos A

cos (A + B)= cos A cos B — sin A sinB

Put A=B

= cos (A + A)= cos A cos A — sin A sin A

= ( cos² A — sin² A ) ---------formula (1)

(sin² A + cos² A = 1, sin² A = 1— cos² A , cos² A = 1 — sin² A )-------------------------------(a)

= cos² A — ( 1— cos² A )

= cos² A — 1 + cos² A

= ( 2 cos² A — 1 )-------------- formula (2)

From (a) above put cos² A = 1 — sin² A in formula (1)

=( 1 — sin² A ) — sin² A

=( 1 — 2 sin² A )----------------formula (3)

∴ cos ( 2A ) = ( cos² A — sin² A ) , ( 2 cos² A — 1 ) & ( 1 — 2 sin² A )

   sin ( 2A )= 2 sin A cos A

 

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K

Karthik | 02/01/2017

Thank you

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