Use the formulas for sin (A +- B) and cos (A +-B) to prove the double angle formulas for sin 2A and cos 2A?

Question

Formulas:

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

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

sin 2A = 2sin (A) cos (A)

cos 2A = 1 - sin^(2) (A)

Answer 1

i) To Prove: sin 2A = 2sin (A) cos (A)

using formula

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

sin(2A)=sin (A + A) = sin (A) cos (A) + sin (A) cos (A)

=>sin (2A) = 2sin(A) cos(A)

hence proved

(ii)To Prove: cos 2A = 1 - 2sin^(2) (A) (actually you have given this wrong)

using formula

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

cos(2A)=cos(A + A) = cos(A) cos (A) - sin (A) sin (A)

cos (2A) = cos^(2) (A) - sin^(2) A

using: sin^(2) A + cos^(2) A =1

cos (2A) = (1 - sin^(2) A - sin^(2) A)

= 1 - 2sin^(2) A

hence proved!!

Answer 2

sin 2A

sin (A + A)

sin A cos A + cos A sin A

sin A cos A + sin A cos A

2 sin A cos A

cos 2A

cos (A + A)

cos A cos A - sin A sin A

cos ² A - sin ² A

1 - 2 sin ² A

Answer 3

  hi  

that 's true

Answer 4

sin(2z)=sin(z)cos(z)+sin(z)cos(z)

=2sin(z)cos(z)

cos(z+z)=cos(z)cos(z)-sin(z)sin(z)

=cos^2(z)-sin^2(z)

=(1-sin^2(z))-sin^2(z)

=1-2sin^2(z)

only takes 2 mins  

Answer 5

(i) Prove: sin 2A = 2sin (A) cos (A)

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

since we want 2A,

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

sin (2A) = 2sin(A)cos(A) as required

(ii) prove: cos 2A = 1 - sin^(2) (A) < this double angle formula is wrong

cos(A +-B) = cos(A) cos (B) -+ sin (A) sin (B) < u got this rule wrong.

since we want 2A as usual

cos(A + A) = cos(A) cos (A) - sin (A) sin (A)

cos (2A) = cos^(2) (A) - sin^(2) A

using: sin^(2) A + cos^(2) A =1

cos (2A) = (1 - sin^(2) A - sin^(2) A)

             = 1 - 2sin^(2) A