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

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### 2 Answers

It is given that a + b = pi/2

=> a = pi/2 - b

sin 2a

=> sin 2*(pi/2 - b)

=> sin 2b

=> sin 2b

sin 2a + sin 2b

=> 2*sin 2b

2*cos (a - b)

=> 2*cos (pi/2 - b - b)

=> 2*cos (pi/2 - 2b)

=> 2*sin 2b

**This proves that sin 2a+sin 2b = 2*cos(a-b) if a+b = pi/2.**

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.**