Answer :
Answer:
Only points on the circle satisfy the given inequality.
Step-by-step explanation:
Given: Unit circle
To find: portion of the unit circle which satisfies the trigonometric inequality [tex]\sin ^2\theta +\cos ^2\theta \geq 1[/tex]
Solution:
In the given figure, OA = 1 unit (as radius of the unit circle equal to 1)
[tex]\sin \theta[/tex] = side opposite to [tex]\theta[/tex]/hypotenuse
[tex]\cos \theta[/tex] = side adjacent to [tex]\theta[/tex]/hypotenuse
[tex]\sin \theta =\frac{AB}{AO}\\\sin \theta =\frac{AB}{1}\\AB=\sin \theta[/tex]
[tex]\cos \theta=\frac{OB}{AO}\\\cos \theta =\frac{OB}{1}\\OB=\cos \theta[/tex]
So, coordinates of A = [tex]\left ( \cos \theta ,\sin \theta \right )[/tex]
For any point (x,y) on the unit circle with centre at origin, equation of circle is given by [tex]x^2+y^2=1[/tex]
Put [tex](x,y)=\left ( \cos \theta ,\sin \theta \right )[/tex]
[tex]\cos ^2\theta +\sin ^2\theta =1[/tex]
So, [tex](x,y)=\left ( \cos \theta ,\sin \theta \right )[/tex] satisfies the equation [tex]x^2+y^2=1[/tex]
For points [tex](x,y)=\left ( \cos \theta ,\sin \theta \right )[/tex] inside the circle, [tex]\cos ^2\theta +\sin ^2\theta <1[/tex]
For points [tex](x,y)=\left ( \cos \theta ,\sin \theta \right )[/tex] outside the circle, [tex]\cos ^2\theta +\sin ^2\theta >1[/tex]
So, only points on the circle satisfy the given inequality.
