Question

The points $O\left( {0,0} \right)$, $A\left( {\cos \alpha ,\sin \alpha } \right)$ and $B\left( {\cos \beta ,\sin \beta } \right)$ are the vertices of a right-angled triangle if
A. $\sin \dfrac{{\alpha - \beta }}{2} = \dfrac{1}{{\sqrt 2 }}$
B. $\cos \dfrac{{\alpha - \beta }}{2} = - \dfrac{1}{{\sqrt 2 }}$
C. $\cos \dfrac{{\alpha - \beta }}{2} = \dfrac{1}{{\sqrt 2 }}$
D. $\sin \dfrac{{\alpha - \beta }}{2} = - \dfrac{1}{{\sqrt 2 }}$

Answer 1

We will first determine the distance between the points of $AB$ and $AC$ using the distance formula, $d = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}}$ and then we will determine whether the distance $AB$ and $AC$are same or not. If yes, then they both are not the hypotenuse of the right-angled triangle. Thus, we get that the side $AB$ is perpendicular on side $BC$. Next we will let the slopes of both sides as ${m_1}$ and ${m_2}$ and as the sides are perpendicular so, the product of slopes will be equal to $- 1$. Now, we will find the values of the slope and then simplify the expression to find the required relation.

Complete step by step answer:

We will first consider the given coordinates $O\left( {0,0} \right)$, $A\left( {\cos \alpha ,\sin \alpha } \right)$ and $B\left( {\cos \beta ,\sin \beta } \right)$ which represent the vertices of the right-angled triangle.

Now, we will find the distance between the coordinates of $A$ and $B$ using the formula $d = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}}$.

$$\Rightarrow d = \sqrt {{{\left( {\cos \alpha - 0} \right)}^2} + {{\left( {\sin \alpha - 0} \right)}^2}} \\\ \Rightarrow d = \sqrt {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \\\ \Rightarrow d = \sqrt 1 \\\ \Rightarrow d = 1 \\\$$

Next, we will find the distance between the coordinates of $A$ and $C$ using the formula $d = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}}$.

$$\Rightarrow d = \sqrt {{{\left( {\cos \beta - 0} \right)}^2} + {{\left( {\sin \beta - 0} \right)}^2}} \\\ \Rightarrow d = \sqrt {{{\cos }^2}\beta + {{\sin }^2}\beta } \\\ \Rightarrow d = \sqrt 1 \\\ \Rightarrow d = 1 \\\$$

As we have got the values for both the lines $AB$ and $AC$ equal thus, $AB$ and $AC$ are not the hypotenuses of the right-angled triangle. So, we obtained the side $BC$ as the hypotenuse of the triangle.
Hence, the side $AB$ is perpendicular to side $AC$ so, we can find the slope of both the lines can be calculated using the formula $m = \dfrac{{\sin x}}{{\cos x}}$ and put the product of the slopes given by ${m_1} \times {m_2} = - 1$.
Thus, the slope of line $AB$ is ${m_1} = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \tan \alpha$ and slope of line $AC$ is ${m_2} = \dfrac{{\sin \beta }}{{\cos \beta }} = \tan \beta$.
Hence, we get,
$\Rightarrow {m_1} \times {m_2} = \tan \alpha \times \tan \beta = - 1$
We can further use the identity that $\tan x = \dfrac{{\sin x}}{{\cos x}}$,
Thus, we have,

$$\Rightarrow \dfrac{{\sin \alpha }}{{\cos \alpha }} \times \dfrac{{\sin \beta }}{{\cos \beta }} = - 1 \\\ \Rightarrow \sin \alpha \sin \beta = - \cos \alpha \cos \beta \\\ \Rightarrow \sin \alpha \sin \beta + \cos \alpha \cos \beta = 0 \\\$$

As we know that the identity $\cos \left( {\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$,
So, we get,

$$\Rightarrow \cos \left( {\alpha - \beta } \right) = 0 \\\ \Rightarrow \alpha - \beta = \pm \dfrac{\pi }{2} \\\$$

Now, we will divide the obtained equation by 2 thus, we get,
$\Rightarrow \dfrac{{\alpha - \beta }}{2} = \pm \dfrac{\pi }{4}$
Now, we will multiply both sides by $\cos$ and we have,

$$\Rightarrow \cos \left( {\dfrac{{\alpha - \beta }}{2}} \right) = \pm \cos \left( {\dfrac{\pi }{4}} \right) \\\ \Rightarrow \cos \left( {\dfrac{{\alpha - \beta }}{2}} \right) = \pm \dfrac{1}{{\sqrt 2 }} \\\$$

Since, the value of $\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$
Hence, option B and C are correct.

Note: It is necessary to determine the side as hypotenuse for further simplification. As the other two lines are perpendicular to each other so, we have to use the property of slope that is ${m_1}{m_2} = - 1$. Remember the identity that ${\cos ^2}x + {\sin ^2}x = 1$ and trigonometric values $\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$. As we have obtained the slopes in terms of $\tan$ so, we can simplify it and convert it into $\sin$ and $\cos$ terms. Remember the trigonometric identity $\cos \left( {\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$. Remember the formula for slope of the line and as the lines are perpendicular, we have to use ${m_1} \times {m_2} = - 1$. Substitution should be done properly.