Question

If sin a = 4/5, then find the value of sin 2a.

Answer 1

You need to apply the double angle formula for the sine:

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

Since sin(A) = 4/5:

sin(2A) = 2*(4/5)*cos(A) = (8/5)cos(A)

To find cos(A), use the Pythagorean identity:

sin2(A) + cos2(A) = 1

cos2(A) = 1 - sin2(A)

cos2(A) = 1 - (4/5)2

cos2(a) = 9/25

√cos2(A) = √(9/25)

cos(A) = ±3/5

So:

sin(2A) = (8/5)*(±3/5) = 24/25

HOPE IT HELPS..

Answer 2

____________________________________________________________

♦ Trigonometric Ratios ♦

→ sin ( A + B ) = [ sin A cos B + sin B cos A ]

 • Substituting 'A' instead, we arrive at :
  → sin ( A + A ) = [ sin A cos A + sin A cos A ]
                           = 2 sin A cos A
____________________________________________________________

→ Given : sin A = ( 4 / 5 )

 • Utilizing cos A = √( 1 - sin² A ), we get :
      → cos A = ( 3 / 5 ) ← 

 • However, the value : sin A = ( 4 / 5 ) is most often used in context with the right triangles.. It's advisable to visualize a Right Triangle for the same .
  [ Refer to the Attachment ]
____________________________________________________________

→ Inserting the Values : 
   • sin 2A = 2 sin A cos A = 2 ( 4 / 5 )( 3 / 5 ) = ( 24 / 25 )
____________________________________________________________

• Proofs for above identity and further such identities can be found on :
  → https://quizlet.com/13179529/trigonometric-identities-for-class-11-flash-cards/

 → http://www.themathpage.com/atrig/sum-proof.htm 
____________________________________________________________

^_^ Hope it helps [ Make sure you get along with the 3-4-5 triangle ]