Solveeit Logo

Question

Question: The number of points of intersection of curve sinx = cosy and x² + y² = 1, is-...

The number of points of intersection of curve sinx = cosy and x² + y² = 1, is-

A

3

B

2

C

0

D

infinitely many

Answer

0

Explanation

Solution

We need to solve the system:

sinx=cosyandx2+y2=1.\sin x = \cos y \quad \text{and} \quad x^2 + y^2 = 1.

Step 1. Rewrite the trigonometric equation

Using the identity cosy=sin(π2y)\cos y = \sin\left(\frac{\pi}{2}-y\right), we have:

sinx=sin(π2y).\sin x = \sin\left(\frac{\pi}{2}-y\right).

This implies:

(i)x=π2y+2πkor(ii)x=π(π2y)+2πk,\text{(i)}\quad x = \frac{\pi}{2} - y + 2\pi k \quad \text{or} \quad \text{(ii)}\quad x = \pi - \left(\frac{\pi}{2} - y\right) + 2\pi k,

where kZk \in \mathbb{Z}.

Simplify (ii):

x=ππ2+y+2πk=π2+y+2πk.x = \pi - \frac{\pi}{2} + y + 2\pi k = \frac{\pi}{2} + y + 2\pi k.

So the two cases are:

Case 1: x=π2y+2πk,Case 2: x=π2+y+2πk.\text{Case 1: } x = \frac{\pi}{2} - y + 2\pi k,\quad \text{Case 2: } x = \frac{\pi}{2} + y + 2\pi k.

Step 2. Substitute in the circle equation

Since the circle x2+y2=1x^2+y^2=1 restricts xx and yy to the interval [1,1][-1, 1], the shifts 2πk2\pi k (for any k0k \neq 0) would lead to values far outside this range. Hence, we consider k=0k=0 only.

  • For Case 1:
x=π2y.x = \frac{\pi}{2} - y.

Substitute into the circle:

(π2y)2+y2=1π24πy+2y2=1.\left(\frac{\pi}{2} - y\right)^2 + y^2 = 1 \quad \Longrightarrow \quad \frac{\pi^2}{4} - \pi y + 2y^2 = 1.

This is a quadratic in yy:

2y2πy+(π241)=0.2y^2 - \pi y + \left(\frac{\pi^2}{4} - 1\right) = 0.

The discriminant DD is:

D=π242(π241)=π28(π241)=π22π2+8=8π2.D = \pi^2 - 4 \cdot 2 \left(\frac{\pi^2}{4} - 1\right) = \pi^2 - 8\left(\frac{\pi^2}{4} - 1\right) = \pi^2 - 2\pi^2 + 8 = 8 - \pi^2.

Since π29.87\pi^2 \approx 9.87, we have D<0D < 0. No real solutions exist.

  • For Case 2:
x=π2+y.x = \frac{\pi}{2} + y.

Substitute into the circle:

(π2+y)2+y2=1π24+πy+2y2=1.\left(\frac{\pi}{2} + y\right)^2 + y^2 = 1 \quad \Longrightarrow \quad \frac{\pi^2}{4} + \pi y + 2y^2 = 1.

The quadratic in yy is:

2y2+πy+(π241)=0.2y^2 + \pi y + \left(\frac{\pi^2}{4} - 1\right) = 0.

Its discriminant is:

D=π242(π241)=8π2,D = \pi^2 - 4 \cdot 2 \left(\frac{\pi^2}{4} - 1\right) = 8 - \pi^2,

which is also negative.

Thus, in both cases, no real yy (and hence no xx) satisfy the equations simultaneously.

Step 3. Conclusion

There are no intersection points between the curves.