Question

Which portion of the unit circle satisfies the trigonometric inequality cos^2theta + sin^2theta is greater than or equal to 1. Assume that theta is the angle made by the positive x-axis and a ray from the origin

Answer 1

Answer:

The right half

Step-by-step explanation:

On the unit circle, cos of theta is the "x" in the (x,y) points. All of the x's are positive in the 1st and 4th quadrants, which is the right side.

Answer 2

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 $\sin ^2\theta +\cos ^2\theta \geq 1$

Solution:

In the given figure, OA = 1 unit (as radius of the unit circle equal to 1)

$\sin \theta$ = side opposite to $\theta$/hypotenuse

$\cos \theta$ = side adjacent to $\theta$/hypotenuse

$\sin \theta =\frac{AB}{AO}\\\sin \theta =\frac{AB}{1}\\AB=\sin \theta$

$\cos \theta=\frac{OB}{AO}\\\cos \theta =\frac{OB}{1}\\OB=\cos \theta$

So, coordinates of A = $\left ( \cos \theta ,\sin \theta \right )$

For any point (x,y) on the unit circle with centre at origin, equation of circle is given by $x^2+y^2=1$

Put $(x,y)=\left ( \cos \theta ,\sin \theta \right )$

$\cos ^2\theta +\sin ^2\theta =1$

So, $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ satisfies the equation $x^2+y^2=1$

For points  $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ inside the circle, $\cos ^2\theta +\sin ^2\theta$

For points  $(x,y)=\left ( \cos \theta ,\sin \theta \right )$ outside the circle, $\cos ^2\theta +\sin ^2\theta >1$

So, only points on the circle satisfy the given inequality.