Question

prove that sin 6x + sin 2x= 2sin 4x cos 2x​

Answer 1

$\maltese$Given to prove that :-

$sin6x + sin2x = 2sin4x\: cos2x$

$\maltese$SOLUTION :-

Take L.H.S that is sin6x +sin2x

It is in form of

$sinC + sinD = 2sin \bigg(\dfrac{C+D}{2} \bigg)cos \bigg(\dfrac{C-D}{2} \bigg)$

So,

$sin6x +sin2x = 2 sin \bigg(\dfrac{6x+2x}{2} \bigg) cos\bigg(\dfrac{6x-2x}{2} \bigg)$

$sin6x+sin2x = 2sin\bigg(\dfrac{8x}{2} \bigg)cos\bigg(\dfrac{4x}{2} \bigg)$

$sin6x + sin2x = 2sin4x cos2x$

${\boxed{sin6x+sin2x = 2sin4x cos2x}}$

$\maltese$${\boxed{Hence\: \: proved!!}}$

$\maltese$Know more formulae :-

$sinC- sinD = 2sin\bigg(\dfrac{C-D}{2} \bigg)cos\bigg(\dfrac{C+D}{2} \bigg)$

$cosC +cosD = 2cos\bigg(\dfrac{C+D}{2} \bigg)cos\bigg(\dfrac{C-D}{2} \bigg)$

$cosC- cosD = -2sin\bigg(\dfrac{C+D}{2} \bigg)sin\bigg(\dfrac{C-D}{2} \bigg)$

$sin(A+B)= sinAcosB + sinBcosA$

$sin(A-B) = sinAcosB- sinBcosA$

$cos(A+B) = cosAcosB - sinAsinB$

$cos(A-B) = cosAcosB + sinAsinB$

$tan(A+B) = \dfrac{tanA+tanB}{1-tanAtanB}$

$tan(A-B) = \dfrac{tanA-tanB}{1+tanAtanB}$

$cot(A+B) = \dfrac{cotB cotA -1}{cotB + cotA}$

$cot(A-B) = \dfrac{cotB cotA + 1}{cotB-cotA}$