Given:
[tex]\theta_1[/tex] is located in IV Quadrant.
[tex]\sin (\theta_1)=-\dfrac{24}{25}[/tex]
To find:
The value of [tex]\cos \theta_1[/tex].
Solution:
We have,
[tex]\sin (\theta_1)=-\dfrac{24}{25}[/tex]
We know that,
[tex]\sin^2 \theta+\cos^2\theta=1[/tex]
So,
[tex]\sin^2 (\theta_1)+\cos^2(\theta_1)=1[/tex]
[tex]\left(-\dfrac{24}{25}\right)^2+\cos^2(\theta_1)=1[/tex]
[tex]\cos^2(\theta_1)=1-\dfrac{576}{625}[/tex]
[tex]\cos^2(\theta_1)=\dfrac{625-576}{625}[/tex]
Taking square root on both sides, we get
[tex]\cos (\theta_1)=\pm \sqrt{\dfrac{49}{625}}[/tex]
[tex]\cos (\theta_1)=\pm \dfrac{7}{25}[/tex]
[tex]\theta_1[/tex] is located in IV Quadrant. In IV Quadrant the value of cos is positive. So,
[tex]\cos (\theta_1)=\dfrac{7}{25}[/tex]
Therefore, [tex]\cos (\theta_1)=\dfrac{7}{25}[/tex].
