Question

If $0<\theta <{45}^o$, then $(\cos \theta -\sin \theta )$ is

• A. always negative
• B. always positive
• C. sometimes positive and sometimes negative
• D. zero

Answer 1

Answer: (B)

always positive

$(\cos \theta -\sin \theta )$
Multiplying and dividing by $\sqrt{2}$
$\sqrt{2}(\frac{\cos \theta }{\sqrt{2}}-\frac{\sin \theta }{\sqrt{2}})$

$=\sqrt{2}(\sin \frac{\pi }{4}\cos \theta -\cos \frac{\pi }{4}\sin \theta )$

$\because \sin \frac{\pi }{4}=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$

$=\sqrt{2}\sin (\frac{\pi }{4}-\theta )$

Since $0<\theta <\frac{\pi }{4}$
$\Rightarrow \frac{\pi }{4}>\frac{\pi }{4}-\theta >0$

Therefore
$\sin (\frac{\pi }{4}-\theta )>0$ in the interval $0<\theta <\frac{\pi }{4}$
Hence,

$(\cos \theta -\sin \theta )$ is positive in the interval $0<\theta <\frac{\pi }{4}$