Question

(sin3x-cos3x/sinx+cosx)+1​

Answer 1

$\textbf{To find:}$

$\text{The value of (\dfrac{sin3x-cos3x}{sinx+cosx})+1 }$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{sin3x-cos3x}{sinx+cosx}+1$

$\text{Using the identity,}$

$\boxed{\bf\,sin3A=3\,sinA-4\,sin^3A}$

$\boxed{\bf\,cos3A=4\,cos^3A-3\,cosA}$

$=\dfrac{(3\,sinx-4\,sin^3x)-(4\,cos^3x-3\,cosx)}{sinx+cosx}+1$

$=\dfrac{3(sinx+cosx)-4(sin^3x+cos^3x)}{sinx+cosx}+1$

$=\dfrac{3(sinx+cosx)-4(sinx+cosx)(sin^2x-sinx\,cosx+cos^2x)}{sinx+cosx}+1$

$=\dfrac{3(sinx+cosx)-4(sinx+cosx)(1-sinx\,cosx)}{sinx+cosx}+1$

$=\dfrac{(sinx+cosx)(3-4(1-sinx\,cosx))}{sinx+cosx}+1$

$=3-4(1-sinx\,cosx)+1$

$=3-4+4\,sinx\,cosx+1$

$=-1+4\,sinx\,cosx+1$

$=4\,sinx\,cosx$

$=2(2\,sinx\,cosx)$

$=2\,sin2x$

$\textbf{Answer:}$

$\boxed{\bf\,(\dfrac{sin3x-cos3x}{sinx+cosx})+1=2\,sin2x}$

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If cos2x-cosx= sin4x - sinx(

where tan doesnot equal to 1) find value of cos3x - sin3x

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