Solveeit Logo
Login

Question

Question: f \( \cos \left( \alpha -\beta \right)=1 \) and \( \cos \left( \alpha +\beta \right)=\dfrac{1}{e} \)...

f cos⁡(α−β)=1\cos \left( \alpha -\beta \right)=1 and cos⁡(α+β)=1e\cos \left( \alpha +\beta \right)=\dfrac{1}{e} , where α,β∈[−π,π]\alpha ,\beta \in \left[ -\pi ,\pi \right] . Then the number of ordered pairs of (α,β)\left( \alpha ,\beta \right) is:
A. 0
B. 1
C. 2
D. 4

Explanation

Solution

Hint : We can solve this question by forming one a relation between α\alpha and β\beta from one of the given trigonometric equations and using that relation, put the value of one variable in terms of the other variable and put it into the other equation and obtain its different values in the given domain. Here, we will use the equation cos⁡(α−β)=1\cos \left( \alpha -\beta \right)=1. This way we will get the number of ordered pairs for (α,β)\left( \alpha ,\beta \right)

Complete step-by-step answer :
Now, it is given that cos⁡(α−β)=1\cos \left( \alpha -\beta \right)=1
⇒α−β=2nπ\Rightarrow \alpha -\beta =2n\pi
Now, since its given in the question that α,β∈[−π,π]\alpha ,\beta \in \left[ -\pi ,\pi \right] ,we can form the following relations:
−π≤α≤π-\pi \le \alpha \le \pi .....(1)
−π≤β≤π ⇒π≥−β≥−π \begin{aligned} & -\pi \le \beta \le \pi \\\ & \Rightarrow \pi \ge -\beta \ge -\pi \\\ \end{aligned}
⇒−π≤−β≤π\Rightarrow -\pi \le -\beta \le \pi .....(2)
Now we will add the inequalities (1) and (2) and as a result we will get:
−2π≤α−β≤2π-2\pi \le \alpha -\beta \le 2\pi
Since we have α−β=2nπ\alpha -\beta =2n\pi
−2π≤2nπ≤2π-2\pi \le 2n\pi \le 2\pi
⇒−1≤n≤1 ⇒n∈−1,0,1 \begin{aligned} & \Rightarrow -1\le n\le 1 \\\ & \Rightarrow n\in \\{-1,0,1\\} \\\ \end{aligned}
Therefore, we get α−β∈−2π,0,2π\alpha -\beta \in \\{-2\pi ,0,2\pi \\}
Now, we can write α\alpha in terms of β\beta as:
⇒α∈β−2π,β,β+2π\Rightarrow \alpha \in \\{\beta -2\pi ,\beta ,\beta +2\pi \\}
Now we can calculate α+β\alpha +\beta as:
⇒α+β∈2β−2π,2β,2β+2π\Rightarrow \alpha +\beta \in \\{2\beta -2\pi ,2\beta ,2\beta +2\pi \\}
Now, cos⁡(α+β)\cos (\alpha +\beta ) can be calculated as:
⇒cos⁡(α+β)=cos⁡(2β−2π),cos⁡(2β),cos⁡(2β+2π)\Rightarrow \cos \left( \alpha +\beta \right)=\\{\cos \left( 2\beta -2\pi \right),\cos \left( 2\beta \right),\cos \left( 2\beta +2\pi \right)\\}
Since cos⁡(2β−2π)=cos⁡2β=cos⁡(2β+2π)\cos \left( 2\beta -2\pi \right)=\cos 2\beta =\cos \left( 2\beta +2\pi \right)
Therefore, cos⁡(α+β)=cos⁡2β\cos \left( \alpha +\beta \right)=\cos 2\beta and α=β\alpha =\beta
Now, it is given that cos⁡(α+β)=1e\cos (\alpha +\beta )=\dfrac{1}{e}
Thus, cos⁡2β=1e\cos 2\beta =\dfrac{1}{e} which is less than 1
Since, β∈[−π,π]⇒2β∈[−2π,2π]\beta \in \left[ -\pi ,\pi \right]\Rightarrow 2\beta \in \left[ -2\pi ,2\pi \right] and also, 1e<1\dfrac{1}{e}<1
Therefore, there are 4 values of β\beta between [−2π,2π]\left[ -2\pi ,2\pi \right] for which it satisfies the given two trigonometric equations as ‘cos’ function has values belonging to [−1,1]\left[ -1,1 \right] and each value is given by any two angles(in radians) belonging to [−0,2π]\left[ -0,2\pi \right] and since 2β∈[−2π,2π]2\beta \in \left[ -2\pi ,2\pi \right],there will be a total of ‘4’ values for β\beta satisfying the given equations.
Now, we have already established that α=β\alpha =\beta
Therefore, for every value of β\beta ,there is one equal value of α\alpha
Thus, there are a total of 4 ordered pairs for (α,β)\left( \alpha ,\beta \right) satisfying the given conditions.
So, the correct answer is “Option D”.

Note : We can also establish a relation between α\alpha and β\beta by using the other given trigonometric equation but it will result in the use of inverse cosine function which will make the calculation really complicated and long which result in a lot of wastage of time. Even after that long and tedious calculation, we may get an answer but it has a very high possibility of being wrong as that complicated calculation will increase the scope for committing mistakes. Hence the equation used above should be used in establishing the relation between α\alpha and β\beta .