Question

If \[3\sin \theta + 5\cos \theta = 5\], then \[5\sin \theta - 3\cos \theta \] is equal to
A. \[4\]
B. \[ \pm 3\]
C. \[5\]
D. None of the above

Answer 1

Hint: We can solve this using simple algebraic identities. We square on both sides of \[3\sin \theta + 5\cos \theta = 5\]. We expand it using algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]. Then on further simplification using Pythagoras relation between sine, cosine and tangent, that is by using \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] we will have the desired result.

Complete step by step answer:
Given
\[3\sin \theta + 5\cos \theta = 5\]
Now squaring on both sides we have,
\[{\left( {3\sin \theta + 5\cos \theta } \right)^2} = {5^2}\]
Now applying the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\], we have
\[{\left( {3\sin \theta } \right)^2} + {\left( {5\cos \theta } \right)^2} + 2\left( {3\sin \theta } \right)\left( {5\cos \theta } \right) = {5^2}\]
\[9{\sin ^2}\theta + 25{\cos ^2}\theta + 30\sin \theta \cos \theta = 25\]
Now from Pythagoras identity we have \[{\sin ^2}\theta + {\cos ^2}\theta = 1\], using this we have \[{\sin ^2}\theta = 1 - {\cos ^2}\theta \] and \[{\cos ^2}\theta = 1 - {\sin ^2}\theta \]. Applying this we have
\[9\left( {1 - {{\cos }^2}\theta } \right) + 25\left( {1 - {{\sin }^2}\theta } \right) + 30\sin \theta \cos \theta = 25\]
Now expanding the brackets we have
\[9 - 9{\cos ^2}\theta + 25 - 25{\sin ^2}\theta + 30\sin \theta \cos \theta = 25\]
Now grouping we have
\[9 + 25 - 9{\cos ^2}\theta - 25{\sin ^2}\theta + 30\sin \theta \cos \theta = 25\]
Or
\[9 + 25 - \left( {9{{\cos }^2}\theta + 25{{\sin }^2}\theta - 30\sin \theta \cos \theta } \right) = 25\]
Now subtracting 25 on both sides of the equation we have
\[9 - \left( {9{{\cos }^2}\theta + 25{{\sin }^2}\theta - 30\sin \theta \cos \theta } \right) = 0\]
Shifting the terms we have
\[\left( {9{{\cos }^2}\theta + 25{{\sin }^2}\theta - 30\sin \theta \cos \theta } \right) = 9\]
If we observe in the left hand side of the equation it is of the form \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\], where \[a = 3\cos \theta \] and \[b = 5\sin \theta \]. Applying this we have
\[{\left( {3\cos \theta - 5\sin \theta } \right)^2} = 9\]
Now taking square root on both sides we have
\[\left( {3\cos \theta - 5\sin \theta } \right) = \pm \sqrt 9 \]
We know that the sure root of 9 is 3. Then
\[ \Rightarrow 3\cos \theta - 5\sin \theta = \pm 3\].
\[ \Rightarrow 5\sin \theta - 3\cos \theta = \pm 3\].
Thus the required answer is \[ \pm 3\].

So, the correct answer is “Option B”.

Note: There are three Pythagorean trigonometric identities and we should remember them all. The identities \[{\sin ^2}\theta + {\cos ^2}\theta = 1\], \[{\tan ^2}\theta + 1 = {\sec ^2}\theta \] and \[1 + {\cot ^2}\theta = {\csc ^2}\theta \]. We apply this according to the given formula. We also know that while shifting a number or term from one side of the equation to the other side the signs will change.