Answer:
[tex]f(x)=-\sqrt{2}sinx[/tex]
[tex]f_{min}=-\sqrt{2}[/tex]
Step-by-step explanation:
Trigonometric Formulas
One of the basic formulas for the trigonometric functions is the sine of a sum of angles:
[tex]sin(x+y)=sinx\cdot cosy+cosx\cdot siny[/tex]
we are required to simplify the formula
[tex]f(x)=sin(x+3\pi /4)+sin(x-3\pi/ 4)[/tex]
Note we have completed the expression to have more sense
Applying the previous formula twice
[tex]f(x)=sin(x+3\pi /4)+sin(x-3\pi/ 4)[/tex]
[tex]f(x)= sinx\cdot cos3\pi / 4+cosx\cdot sin3\pi/ 4+sinx\cdot cos3\pi/ 4-cosx\cdot sin3\pi / 4[/tex]
Simplifying
[tex]f(x)=2sinx\cdot cos 3\pi /4=-\sqrt{2}sinx[/tex]
[tex]f(x)=-\sqrt{2}sinx[/tex]
The minimum value occurs when the sine is at maximum value, i.e. when sinx=1
[tex]\boxed{f_{min}=-\sqrt{2}}[/tex]
After simplification, We get [tex]f(x)=-\sqrt{2}*sinx[/tex] and minimum value of function is [tex]-\sqrt{2}[/tex]
The given function is,
[tex]f(x)=sin(x+\frac{3\pi}{4} )+sin(x-\frac{3\pi}{4} )[/tex]
We know that,
[tex]sin(A+B)=sin(A)*cos(B)+cos(A)*sin(B)[/tex]
By using above formula. given function can be written as,
[tex]f(x)=sinx*cos\frac{3\pi}{4}+ sinx*cos\frac{3\pi}{4}\\\\f(x)=2sinx*cos\frac{3\pi}{4} \\\\f(x)=-\sqrt{2} *sinx[/tex]
To find minimum value of above function substitute [tex]sinx=1[/tex]
Minimum value of [tex]f(x)=-\sqrt{2}[/tex]
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