Given:
[tex]\cos \theta =\dfrac{3}{5}[/tex]
[tex]\sin \theta <0[/tex]
To find:
The quadrant of the terminal side of [tex]\theta[/tex] and find the value of [tex]\sin\theta[/tex].
Solution:
We know that,
In Quadrant I, all trigonometric ratios are positive.
In Quadrant II: Only sin and cosec are positive.
In Quadrant III: Only tan and cot are positive.
In Quadrant IV: Only cos and sec are positive.
It is given that,
[tex]\cos \theta =\dfrac{3}{5}[/tex]
[tex]\sin \theta <0[/tex]
Here cos is positive and sine is negative. So, [tex]\theta [/tex] must be lies in Quadrant IV.
We know that,
[tex]\sin^2\theta +\cos^2\theta =1[/tex]
[tex]\sin^2\theta=1-\cos^2\theta[/tex]
[tex]\sin \theta=\pm \sqrt{1-\cos^2\theta}[/tex]
It is only negative because [tex]\theta [/tex] lies in Quadrant IV. So,
[tex]\sin \theta=-\sqrt{1-\cos^2\theta}[/tex]
After substituting [tex]\cos \theta =\dfrac{3}{5}[/tex], we get
[tex]\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}[/tex]
[tex]\sin \theta=-\sqrt{1-\dfrac{9}{25}}[/tex]
[tex]\sin \theta=-\sqrt{\dfrac{25-9}{25}}[/tex]
[tex]\sin \theta=-\sqrt{\dfrac{16}{25}}[/tex]
[tex]\sin \theta=-\dfrac{4}{5}[/tex]
Therefore, the correct option is B.
