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Question: Eliminate \( \theta \) between \( a = \cos \left( {\theta - \alpha } \right) \) and \( b = \sin \lef...

Eliminate θ\theta between a=cos⁡(θ−α)a = \cos \left( {\theta - \alpha } \right) and b=sin⁡(θ−β)b = \sin \left( {\theta - \beta } \right) .

Explanation

Solution

Hint : In order to take out the value of θ\theta , firstly multiply cosine inverse both the sides on the first equation, which would lead to opening of the parenthesis and we would get one new equation. Similarly multiplying sine inverse on both the sides of the second equation to open the parenthesis. Then subtract one equation from the other which would lead to cancellation of θ\theta , and our value will be obtained.

Complete step-by-step answer :
We are given with two equations, marking them as (1) and (2):
a=cos⁡(θ−α)a = \cos \left( {\theta - \alpha } \right) ……….(1)
b=sin⁡(θ−β)b = \sin \left( {\theta - \beta } \right) ……….(2)
Now, in (1) multiplying both the sides with cosine inverse (cos⁡−1)\left( {{{\cos }^{ - 1}}} \right) , as there is cosine value given:
a=cos⁡(θ−α) =>cos⁡−1(a)=cos⁡−1(cos⁡(θ−α))   a = \cos \left( {\theta - \alpha } \right) \\\ = > {\cos ^{ - 1}}\left( a \right) = {\cos ^{ - 1}}\left( {\cos \left( {\theta - \alpha } \right)} \right) \;
Since, we know that cos⁡−1(cos⁡θ)=θ{\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta , applying this in above equation in order to cancel out the cosine terms and we get:
cos⁡−1(a)=cos⁡−1(cos⁡(θ−α)) =>cos⁡−1(a)=θ−α   {\cos ^{ - 1}}\left( a \right) = {\cos ^{ - 1}}\left( {\cos \left( {\theta - \alpha } \right)} \right) \\\ = > {\cos ^{ - 1}}\left( a \right) = \theta - \alpha \;
Marking the equation cos⁡−1(a)=θ−α{\cos ^{ - 1}}\left( a \right) = \theta - \alpha as (3).
Similarly, multiplying both sides with sine inverse (sin⁡−1)\left( {{{\sin }^{ - 1}}} \right) in (2) as sine value is given in the equation, and using the same concept of cos⁡−1(cos⁡θ)=θ{\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta for sine, in order to cancel out sine terms, we get:
b=sin⁡(θ−β) =>sin⁡−1(b)=sin⁡−1(sin⁡(θ−β)) =>sin⁡−1(b)=θ−β   b = \sin \left( {\theta - \beta } \right) \\\ = > {\sin ^{ - 1}}\left( b \right) = {\sin ^{ - 1}}\left( {\sin \left( {\theta - \beta } \right)} \right) \\\ = > {\sin ^{ - 1}}\left( b \right) = \theta - \beta \;
Marking the equation sin⁡−1(b)=θ−β{\sin ^{ - 1}}\left( b \right) = \theta - \beta as (4):
Now, subtracting (4) from (3) and we get:
(4)−(3) cos⁡−1(a)−sin⁡−1(b)=(θ−α)−(θ−β)   \left( 4 \right) - \left( 3 \right) \\\ {\cos ^{ - 1}}\left( a \right) - {\sin ^{ - 1}}\left( b \right) = \left( {\theta - \alpha } \right) - \left( {\theta - \beta } \right) \;
Opening up the parenthesis for the above equation:
cos⁡−1(a)−sin⁡−1(b)=(θ−α)−(θ−β) =>cos⁡−1(a)−sin⁡−1(b)=θ−α−θ+β   {\cos ^{ - 1}}\left( a \right) - {\sin ^{ - 1}}\left( b \right) = \left( {\theta - \alpha } \right) - \left( {\theta - \beta } \right) \\\ = > {\cos ^{ - 1}}\left( a \right) - {\sin ^{ - 1}}\left( b \right) = \theta - \alpha - \theta + \beta \;
We can see that θ\theta can be cancelled out, so on further solving, we get that:
cos⁡−1(a)−sin⁡−1(b)=−α+β{\cos ^{ - 1}}\left( a \right) - {\sin ^{ - 1}}\left( b \right) = - \alpha + \beta
Arranging the above equation:
cos⁡−1(a)−sin⁡−1(b)=β−α{\cos ^{ - 1}}\left( a \right) - {\sin ^{ - 1}}\left( b \right) = \beta - \alpha
And, θ\theta is cancelled out between the two equations a=cos⁡(θ−α)a = \cos \left( {\theta - \alpha } \right) and b=sin⁡(θ−β)b = \sin \left( {\theta - \beta } \right) .

Note : Do not apply any formula, which is not known to open the brackets otherwise would lead to confusion in solving the above equations.
Do not make mistakes by directly subtracting the terms in order to cancel out inner terms. It’s an inappropriate step.