Question

The distance between the points costheta,sintheta and sintheta -cos theta is

Answer 1

anwser:

We know the formula for distance between two points which is given by

d=sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

d=sqrt{(sintheta-costheta)^2+(-costheta-sintheta)^2}

d=sqrt{sin^2theta+cos^2theta-2sintheta costheta+cos^2theta+sin^2theta+2sintheta costheta}

d=sqrt{2(sin^2theta+cos^2theta)}d=sqrt{2}

Answer 2

The distance between the given points = $\sqrt{2}$ units

Step-by-step explanation:

Here, $x_{1} =\cos \theta,y_{1} =\sin \theta$ and $x_{2} =\sin \theta,y_{2} =-\cos \theta$

To find, the distance between the given points = ?

We know that,

The distance between the two points

= $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

= $\sqrt{(\sin \theta-\cos \theta)^2+(-\cos \theta-\sin \theta)^2}$

= $\sqrt{(\sin \theta-\cos \theta)^2+(\cos \theta+\sin \theta)^2}$

= $\sqrt{\sin^2 \theta+\cos^2 \theta-2\sin \theta\cos \theta+\sin^2 \theta+\cos^2 \theta+2\sin \theta\cos \theta}$

= $\sqrt{\sin^2 \theta+\cos^2 \theta+\sin^2 \theta+\cos^2 \theta}$

= $\sqrt{2\sin^2 \theta+2\cos^2 \theta}$

= $\sqrt{2(\sin^2 \theta+\cos^2 \theta)}$

Using the trigonometric identity,

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

= $\sqrt{2(1)}$

= $\sqrt{2}$ units

∴ The distance between the given points = $\sqrt{2}$ units