Question

If \csc \theta =\sqrt{2}cscθ= 2 ​ and the reference angle of \thetaθ is 45^\circ 45 ∘ , find both angles in degrees from 0^\circ \theta < 360^{\circ}0 ∘ ≤θ<360 ∘ and both angles in radians from 0 \le \theta < 2\pi.0≤θ<2π

Answer 1

Answer:

$45^\circ$ and $135^\circ$ are the values of θ in degree.

$\frac{\pi}{4}$ and $\frac{3\pi}{4}$ are the values of θ in radian are $\frac{\pi}{4}$ and $\frac{3\pi}{4}$.

Step-by-step explanation:

Given that,

$csc\theta = \sqrt2$

$\Rightarrow csc\theta = csc 45^\circ$

$\Rightarrow \theta = 45^\circ$

Degree to radian,

$180^\circ = \pi$

or, $1^\circ =\frac{\pi}{180}$

csc is reciprocal function of sin.

Sin is positive only first co-ordinate and second co-ordinate.

So, csc is also positive only first co-ordinate and second co-ordinate.

Since $0^\circ\leq \theta\leq 360^\circ$

Therefore

$csc\theta = \sqrt2$

$\Rightarrow csc\theta = csc (45^\circ) \ or\ csc (90^\circ+45^\circ)$

$\Rightarrow \theta =45^\circ , 135^\circ$

$\Rightarrow \theta =\frac{\pi \times 45}{180} , \frac{\pi \times 135}{180}$

$\Rightarrow \theta =\frac{\pi}{4} , \frac{3\pi}{4}$

$45^\circ$ and $135^\circ$ are the values of θ in degree.

$\frac{\pi}{4}$ and $\frac{3\pi}{4}$ are the values of θ in radian are $\frac{\pi}{4}$ and $\frac{3\pi}{4}$.