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Grade: 11
        How to find range of sinx-cosx
3 years ago

Answers : (4)

radhika
11 Points
							
asinx – bcosx
min value = -√a2+b= -√1+1 = -√2
max value = √a2+b2 = √2

 
3 years ago
AUREA
61 Points
							
f(x) = sinx – cosx
     =√2 x 1/√2 ( sinx – cosx)
     =√2 x { (1/√2)sinx – (1/√2)cosx }
     =√2 sin (x – pi/4)
range of sine function is [-1,1]
so range of f(x) will be [-√2,√2]
3 years ago
grenade
2056 Points
							
The max value equals root2 and minimum minus root 2
Thence the range is between min and maz
Hence the answer would be from minus root 2to root 2
3 years ago
ng29
1698 Points
							
just find the max and min values of this equation by differentiating it 
we get min = -(2)1/2  and max =(2)1/2
so its range of function 
 
approve if useful
3 years ago
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