# Given that a+b = pi/2, prove that sin 2a+sin 2b=2cos(a-b)?

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### 2 Answers

We have a + b = pi/2

=> a = pi/2 - b

=> 2a = pi - 2b

Now sin 2a + sin 2b

=> sin 2a + sin (pi - 2b)

As sin ( pi - x) = sin x

=> sin 2a + sin 2b

=> 2 sin a cos a + 2 sin b cos b ...(1)

Now 2 cos (a - b) = cos a cos b + sin a sin b

=> 2cos a cos b + 2sin a sin b ...(2)

**Therefore, from (1) and (2) we see that the right hand side is equal to the left hand side. **

**Hence sin 2a + sin 2b = 2cos (a-b)**

We'll transform the sum sin 2a+sin 2b into a product:

sin 2a+sin 2b = 2sin [(2a+2b)/2]*cos [(2a-2b)/2]

We'll factorize by 2 inside the brackets:

sin 2a+sin 2b = 2sin [2(a+b)/2]*cos [2(a-b)/2]

We'll simplify and we'll get:

sin 2a+sin 2b = 2sin [(a+b)]*cos [(a-b)]

But, from enunciation, a+b = pi/2.

sin 2a+sin 2b = 2sin [(pi/2)]*cos [(a-b)]

But sin pi/2 = 1

sin 2a+sin 2b = 2cos [(a-b)] q.e.d.