Question

Sin(x+pi/4)-sin(x-pi/4)=1 solve the equation

Answer 1

Sin(α+β)=sin(α)cos(β)+cos(α)sin(β)
sin(α-β)=sin(α)cos(β)-cos(α)sin(β)


$sin(x+ \frac{ \pi }{4} )=sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} \\ sin(x- \frac{ \pi }{4} )=sin(x)cos\frac{ \pi }{4} -cos(x)sin\frac{ \pi }{4} \\ \\ \\sin(x+ \frac{ \pi }{4} )-sin(x- \frac{ \pi }{4} ) =1 \\ sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} -(sin(x)cos\frac{ \pi }{4} -cos(x)sin\frac{ \pi }{4} )=1 \\ sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} -sin(x)cos\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} =1 \\ cos(x)sin\frac{ \pi }{4} +cos(x)sin\frac{ \pi }{4} =1$
$2cos(x)sin\frac{ \pi }{4} =1 \\ sin\frac{ \pi }{4} = \frac{ \sqrt{2} }{2} \\ 2cos(x) \frac{ \sqrt{2} }{2} =1 \\ 2 \cdot \frac{ \sqrt{2} }{2}cos(x) =1 \\ \sqrt{2} cos(x)=1$
$cos(x)= \frac{1}{ \sqrt{2} } \\ x=\pm arccos \frac{1}{ \sqrt{2} }+2 \pi k , k \in Z \\ x=\pm \frac{ \pi }{4} +2 \pi k , k \in Z$