Question

sin(a+b)cos(a-b) + sin(a-b)cos(a+b)=sin2a​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: sin(a + b)cos(a - b) + sin(a - b)cos(a + b)$

On multiply and divide by 2, we get

$\rm \: \dfrac{1}{2}\bigg(sin(a + b)cos(a - b) + sin(a - b)cos(a + b)\bigg)$

We know,

$\boxed{\tt{ 2sinxcosy = sin(x + y) + sin(x - y) \: }}$

So, using this identity, we get

$\rm \: = \dfrac{1}{2}\bigg(sin(a + b + a - b) + sin(a + b - a + b) + sin(a - b + a + b) + sin(a - b - a - b)\bigg)$

$\rm \: = \dfrac{1}{2}\bigg(sin2a + sin2b + sin2a + sin( - 2b)\bigg)$

We know,

$\boxed{\tt{ sin( - x) = - sinx \: }}$

So, using this, we get

$\rm \: = \dfrac{1}{2}\bigg(sin2a + sin2b + sin2a - sin2b\bigg)$

$\rm \: = \dfrac{1}{2}\bigg(2 \: sin2a \bigg)$

$\rm \: = \: sin2a$

Hence,

$\boxed{\tt{ sin(a + b)cos(a - b) + sin(a - b)cos(a + b) = sin2a}}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ cos(x + y) + cos(x - y) = 2cosxcosy \: }}$

$\boxed{\tt{ cos(x - y) - cos(x + y) = 2sinx \: siny \: }}$

$\boxed{\tt{ sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }}$

$\boxed{\tt{ sinx - siny = 2sin\bigg[\dfrac{x - y}{2} \bigg]cos\bigg[\dfrac{x + y}{2} \bigg] \: }}$

$\boxed{\tt{ cosx + cosy = 2cos\bigg[\dfrac{x - y}{2} \bigg]cos\bigg[\dfrac{x + y}{2} \bigg] \: }}$

$\boxed{\tt{ cosx - cosy = - 2sin\bigg[\dfrac{x - y}{2} \bigg]sin\bigg[\dfrac{x + y}{2} \bigg] \: }}$

Answer 2

Step-by-step explanation:

$\boxed{\sf \red{ \underline{\underline{\mathbb {GIVEN:}}} \sf \sin(A+B) \cos(A-B) + \sin(A-B) \cos(A+B)= \sin2A}}$

$\boxed{ \sf \pink{ \underline {\underline{\bullet To \: \: Find : }} \: \sin(2A) }}$

$\sf \blue{ \implies\sin(A+B) \cos(A-B) + \sin(A-B) \cos(A+B)}$

$\sf \purple{ \implies\sin(A+ \cancel{B }+ A - \cancel{B}) }$

$\sf \orange { \implies\sin(2A) = L.H.S}$