Question

What is the range of sin x- cos x
Please don't tell the range of sin x+cosx

Answer 1

f(x) = sinx – cosx
=√2 x 1/√2 ( sinx – cosx)
=√2 x { (1/√2)sinx – (1/√2)cosx }
=√2 sin (x – π/4)
range of sine function is [-1,1]
so range of f(x) will be [-√2,√2]

Answer 2

Range = $[-\sqrt{2} ,\sqrt{2}]$

Step-by-step explanation:

Let $f(x)=\sin x -\cos x$

$(\sin x-\cos x)^{2} =\sin ^{2} x+\cos ^{2} x-2\sin x\cos\x$

⇒$(\sin x-\cos x)^{2} =1-2\sin x\cos\x$

[∵ $\sin ^{2} A+\cos ^{2} A=1$]

⇒$(\sin x-\cos x)^{2} =1-\sin 2x$

[ ∵$\sin 2x=2\sin x\cos\x$]

⇒ $(\sin x-\cos x)^{2} =1-1=0$

[ ∵$\sin 2x$ is maximum is 1

⇒ $\sin x=\cos x=$

∴ Range = [- $\sqrt{2} ,\sqrt{2}$]