Question

If $-\frac{\pi }{4}<\theta <\frac{\pi }{4}$,then the value of $cos\theta -sin\theta$ is

• A. always negative
• B. always positive
• C. sometimes positive, sometimes negative
• D. 0

Answer 1

Answer: (B)

always positive

$(cos\theta -sin\theta )$
Multiplying and dividing by $\sqrt{2}$
Therefore
$\sqrt{2}(\frac{cos\theta }{\sqrt{2}}-\frac{sin\theta }{\sqrt{2}})$
$=\sqrt{2}(sin\frac{\pi }{4}cos\theta -cos\frac{\pi }{4}sin\theta )$ ...$(sin\frac{\pi }{4}=cos\frac{\pi }{4}=\frac{1}{\sqrt{2}})$
$=\sqrt{2}sin(\frac{\pi }{4}-\theta )$
Since $\frac{-\pi }{4}<\theta <\frac{\pi }{4}$
Therefore minimum value is $0$ at $\theta =\frac{\pi }{4}$
$sin(\frac{\pi }{4}-\theta )>or=0$ in the interval $\frac{-\pi }{4}<\theta <\frac{\pi }{4}$
Hence $(cos\theta -sin\theta )$ is positive in the interval $\frac{-\pi }{4}<\theta <\frac{\pi }{4}$