Grade 11Trigonometry
How to find range of sinx-cosx
4 Answers
radhika
asinx – bcosx
min value = -√a2+b2 = -√1+1 = -√2
max value = √a2+b2 = √2
min value = -√a2+b2 = -√1+1 = -√2
max value = √a2+b2 = √2
AUREA
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]
grenade
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
ng29
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
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