Question

If $5\sin x + 4\cos x = 3$, then find the value of $4\sin x - 5\cos x = $ ?
A. $4$
B. $4\sqrt 2 $
C. $3\sqrt 2 $
D. $\sqrt 2 $

Answer 1

Hint: Initially, we will do squaring on both left-hand side (L.H.S) and right-hand side (R.H.S) of the given expression that is given by:
 $5\sin x + 4\cos x = 3$
After that we will solve it by using the formula of algebraic identities which is square of the addition of two variables. Then, we will assume some variable like ‘k’ on the right-hand side (R.H.S) of the given equation which has to be found. Then, similarly we will do squaring on the both sides of the equation given by:
 $4\sin x - 5\cos x = ? $
After that, again we will apply the formula of algebraic identities which is square of the difference of the two variables, in order to expand it. Finally, after expanding both the given equations, we will do the addition of both the simplified expression to get the result.

Formula used: ${(x + y)^2} = {x^2} + 2xy + {y^2}$
 ${(x - y)^2} = {x^2} - 2xy + {y^2}$

Complete step-by-step solution:
As per the question, the expression is given as:
$5\sin x + 4\cos x = 3$ ----- (1)
Squaring on the both (L.H.S and R.H.S) side of equation (1), we get:
 ${(5\sin x + 4\cos x)^2} = {(3)^2}$
Now, apply the algebraic identities formula of ${(x + y)^2}$ in the above equation, we get:
 ${(5\sin x)^2} + 2(5\sin x)(4\cos x) + {(4\cos x)^2} = 9$
$ \Rightarrow 25{\sin ^2}x + 40 \cdot \sin x \cdot \cos x + 16{\cos ^2}x = 9$ ---- (2)
Similarly, the value of equation that has to be found is given by:
$4\sin x - 5\cos x = ? $------ (3)
Now, let the R.H.S side of equation (3) be ’k’.
$4\sin x - 5\cos x = k $ ----- (4)
Now, squaring on the both (L.H.S and R.H.S) side of equation (4), we get:
 ${(4\sin x - 5\cos x)^2} = {(k)^2}$
Now, apply the algebraic identities formula of ${(x - y)^2}$ in the above equation, we get:
${(4\sin x)^2} - 2 \cdot (4\sin x) \cdot (5\cos x) + {(5\cos x)^2} = {k^2}$
$ \Rightarrow 16{\sin ^2}x - 40 \cdot \sin x \cdot \cos x + 25{\cos ^2}x = {k^2} $ ------ (5)
Now, adding both sides of equation (2) and (5), we get:
 $(25{\sin ^2}x + 40 \cdot \sin x \cdot \cos x + 16{\cos ^2}x) + (16{\sin ^2}x - 40 \cdot \sin x \cdot \cos x + 25{\cos ^2}x) = 9 + {k^2} $
 $ \Rightarrow (25{\sin ^2}x + 16{\sin ^2}x) + (40 \cdot \sin x \cdot \cos x - 40 \cdot \sin x \cdot \cos x) + (16{\cos ^2}x + 25{\cos ^2}x) = 9 + {k^2} $
$ \Rightarrow 41{\sin ^2}x + 41{\cos ^2}x = 9 + {k^2} $
$ \Rightarrow 41({\sin ^2}x + {\cos ^2}x) = 9 + {k^2} $
$ \Rightarrow 41(1) = 9 + {k^2} $
$ \Rightarrow {k^2} = 41 - 9 = 32 $
$ \Rightarrow k = \sqrt {32} $
\[ \Rightarrow k = 4\sqrt 2 \]
Thus, the value of $4\sin x - 5\cos x $ is $4\sqrt 2 $.

Hence, the correct option is B.

Note: Always, in such type of questions ( that is whenever some trigonometric expression is given or any variable given as different type, along with their equation value and we have to find out the solution of some other expression but with the same variable) try to first do the squaring of both the given expressions and apply the relevant algebraic identities formula.
Generally, try to remember out the basic two algebraic identities formula:
 ${(x + y)^2} = {x^2} + 2xy + {y^2}$
 ${(x - y)^2} = {x^2} - 2xy + {y^2}$