Question

If $a\cos \theta - b\sin \theta = c$ then show that $a\sin \theta + b\cos \theta = \pm \sqrt {{a^2} + {b^2} - {c^2}} $

Answer 1

Hint: Here we will convert the expression into a quadratic equation by applying the square and then by using the appropriate algebraic identities $a\sin \theta + b\cos \theta = \pm \sqrt {{a^2} + {b^2} - {c^2}} $ is proved.

Complete step-by-step answer:

Given condition
    $a\cos \theta - b\sin \theta = c$
Let us square on both sides, we get
${\left( {a\cos \theta - b\sin \theta } \right)^2} = {c^2}$
  Now on applying ${(a - b)^2} = {a^2} + {b^2} - 2ab$ formula we get
$ \Rightarrow {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta - 2ab\sin \theta \cos \theta = {c^2}$
We know that ${\cos ^2}\theta = 1 - {\sin ^2}\theta $ and ${\sin ^2}\theta = 1 - {\cos ^2}\theta $
On the applying these value to the above equation we get
$ \Rightarrow {a^2}(1 - {\sin ^2}\theta ) + {b^2}\left( {1 - {{\cos }^2}\theta } \right) - 2ab\cos \theta \sin \theta = {c^2}$
$ \Rightarrow {a^2} - {a^2}{\sin ^2}\theta + {b^2} - {b^2}{\cos ^2}\theta - 2ab\sin \theta \cos \theta = {c^2}$
After rearranging the terms we get
$ \Rightarrow {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta + 2ab\cos \theta \sin \theta = {a^2} + {b^2} - {c^2}$
Let us use ${(a + b)^2}$ formula to rewrite the LHS part, where${(a + b)^2} = {a^2} + {b^2} + 2ab$
On rewriting we get the equation as
$ \Rightarrow {(a\sin \theta + b\cos \theta )^2} = {a^2} + {b^2} - {c^2}$
$ \Rightarrow $$(a\sin \theta + b\cos \theta ) = \pm \sqrt {{a^2} + {b^2} - {c^2}} $
Therefor we showed that $a\sin \theta + b\cos \theta = \pm \sqrt {{a^2} + {b^2} - {c^2}} $
Hence LHS=RHS has been proved.

Note: Use the conversions like ${\cos ^2}\theta = 1 - {\sin ^2}\theta $ and ${\sin ^2}\theta = 1 - {\cos ^2}\theta $ to get the proper answer. Make use of all possible formulas to get the answer in a simple way. Students must take care while using the appropriate identities.