Question

Let $\alpha ,\,\,\beta$ be such that $\pi <(\alpha -\beta )<3\pi$. If $\sin (\alpha )+\sin (\beta )=\dfrac{-21}{65}$ and $\cos (\alpha )+\cos (\beta )=\dfrac{-27}{65}$, then the value of $\cos \left( \dfrac{\alpha -\beta }{2} \right)$ is:
A) $\dfrac{-6}{65}$
B) $\dfrac{3}{\sqrt{130}}$
C) $\dfrac{6}{65}$
D) $\dfrac{-3}{\sqrt{130}}$

Answer 1

According to the given question there are two terms $\alpha$ and $\beta$ is given such that $\pi <(\alpha -\beta )<3\pi$ and there are two equations given. When you solve the two equation by adding and squaring and applying proper trigonometry property that is this particular problem required formula is $\cos (\alpha -\beta )=\cos (\alpha )\cos (\beta )+\sin (\alpha )\sin (\beta )$ we get the value of $\cos \left( \dfrac{\alpha -\beta }{2} \right)$.

Complete step by step answer:
According to the given condition is that $\pi <(\alpha -\beta )<3\pi$
And there are two equation is given that is
$\sin (\alpha )+\sin (\beta )=\dfrac{-21}{65}---(1)$
$\cos (\alpha )+\cos (\beta )=\dfrac{-27}{65}---(2)$
So, we have to find in this question is that we have to find the value of$\cos \left( \dfrac{\alpha -\beta }{2} \right)$
For that first of all we have to simplify the two equation (1) and (2)
To simplify this two equation we have to first of all squaring the equation (1) and equation (2)
By squaring the equation (1)
${{\left( \sin (\alpha )+\sin (\beta ) \right)}^{2}}={{\left( \dfrac{-21}{65} \right)}^{2}}$
By simplifying this equation by using the property of ${{(a+b)}^{2}}={{a}^{2}}+ab+{{b}^{2}}$we get:
${{\sin }^{2}}(\alpha )+{{\sin }^{2}}(\beta )+2\sin (\alpha )\sin (\beta )=\dfrac{{{(21)}^{2}}}{{{(65)}^{2}}}$
By simplifying this,
${{\sin }^{2}}(\alpha )+{{\sin }^{2}}(\beta )+2\sin (\alpha )\sin (\beta )=\dfrac{441}{{{(65)}^{2}}}--(3)$
Similarly, we can also perform for equation (2).
By squaring the equation (2)
${{\left( \cos (\alpha )+\cos (\beta ) \right)}^{2}}={{\left( \dfrac{-27}{65} \right)}^{2}}$
By simplifying this equation by using the property of ${{(a+b)}^{2}}={{a}^{2}}+ab+{{b}^{2}}$we get:
${{\cos }^{2}}(\alpha )+{{\cos }^{2}}(\beta )+2\cos (\alpha )\cos (\beta )=\dfrac{{{(27)}^{2}}}{{{(65)}^{2}}}$
By simplifying this,
${{\cos }^{2}}(\alpha )+{{\cos }^{2}}(\beta )+2\cos (\alpha )\cos (\beta )=\dfrac{729}{{{(65)}^{2}}}--(4)$
Add this equation (3) and equation (4)
${{\sin }^{2}}(\alpha )+{{\sin }^{2}}(\beta )+2\sin (\alpha )\sin (\beta )+{{\cos }^{2}}(\alpha )+{{\cos }^{2}}(\beta )+2\cos (\alpha )\cos (\beta )=\dfrac{441}{{{(65)}^{2}}}+\dfrac{729}{{{(65)}^{2}}}$
By arranging the term we get:
${{\sin }^{2}}(\alpha )+{{\cos }^{2}}(\alpha )+{{\sin }^{2}}(\beta )+{{\cos }^{2}}(\beta )+2\sin (\alpha )\sin (\beta )+2\cos (\alpha )\cos (\beta )=\dfrac{441}{{{(65)}^{2}}}+\dfrac{729}{{{(65)}^{2}}}$
By using the property ${{\sin }^{2}}(\theta )+{{\cos }^{2}}(\theta )=1$and simplify it
$1+1+2\sin (\alpha )\sin (\beta )+2\cos (\alpha )\cos (\beta )=\dfrac{1170}{{{(65)}^{2}}}$
$2+2\sin (\alpha )\sin (\beta )+2\cos (\alpha )\cos (\beta )=\dfrac{1170}{{{(65)}^{2}}}$
Take the 2 common on LHS and use the trigonometric property that is$\cos (\alpha -\beta )=\cos (\alpha )\cos (\beta )+\sin (\alpha )\sin (\beta )$
$2(1+\cos (\alpha )\cos (\beta )+\sin (\alpha )\sin (\beta ))=\dfrac{1170}{{{(65)}^{2}}}$
$2\left( 1+\cos \left( {\alpha -\beta } \right) \right)=\dfrac{1170}{{{(65)}^{2}}}$
This can be reduced using the formula $\cos 2x = 2 \cos x - 1 \Rightarrow 1+\cos 2x = 2 cos x$, so
$4{{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)=\dfrac{1170}{{{(65)}^{2}}}$
By simplifying further we get:
${{\cos }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)=\dfrac{1170}{{{(65)}^{2}}\times 4}$
Take root on both sides
$\cos \left( \dfrac{\alpha -\beta }{2} \right)=\pm \sqrt{\dfrac{130\times 9}{65\times 65\times 4}}$
By solving further we get:
$\cos \left( \dfrac{\alpha -\beta }{2} \right)=\pm \sqrt{\dfrac{130\times {{(3)}^{2}}}{{{(65)}^{2}}\times {{(2)}^{2}}}}$
$\cos \left( \dfrac{\alpha -\beta }{2} \right)=\pm \dfrac{3\times \sqrt{130}}{130}$
By modifying this equation we get:
$\cos \left( \dfrac{\alpha -\beta }{2} \right)=\pm \dfrac{3}{\sqrt{130}}$
Since, $\pi <(\alpha -\beta )<3\pi$
If we divide by 2 on both side then
$\dfrac{\pi }{2}<\dfrac{\alpha -\beta }{2}<\dfrac{3\pi }{2}$ which lies in 2nd and third quadrants. So, Cos function value is negative.
Therefore, $\cos \left( \dfrac{\alpha -\beta }{2} \right)=\dfrac{-3}{\sqrt{130}}$
So, the correct option is option (D).

Note:
Always remember that while solving this type of particular problem use the proper trigonometry property to get the desired value. The condition which is given that is $\pi <(\alpha -\beta )<3\pi$ important (Don’t ignore this) because value which we get that is $\cos \left( \dfrac{\alpha -\beta }{2} \right)$ both positive as well as negative. In this case we consider positive because values is given between $\pi$ and $3\pi$ that’s why students may confuse in that if we get two values then the condition which is given in that that is $\pi <(\alpha -\beta )<3\pi$ Then we get the desired answer. So, the above solution is referred for such types of problems.