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Question: Given \(3\sin \beta +5\cos \beta =5\), then the value of \({{\left( 3\cos \beta -5\sin \beta \right)...

Given 3sin⁡β+5cos⁡β=53\sin \beta +5\cos \beta =5, then the value of (3cos⁡β−5sin⁡β)2{{\left( 3\cos \beta -5\sin \beta \right)}^{2}} .
(a) 9
(b) 95\dfrac{9}{5}
(c) 13\dfrac{1}{3}
(d) 19\dfrac{1}{9}

Explanation

Solution

To solve this question, we will first square both sides of the equation 3sin⁡β+5cos⁡β=53\sin \beta +5\cos \beta =5 and then simplify it using the identity (a+b)2=a2+b2+2ac{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ac. We will also expand the (3cos⁡β−5sin⁡β)2{{\left( 3\cos \beta -5\sin \beta \right)}^{2}} using the identity (a−b)2=a2+b2−2ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab. Once we get the two equations, we will replace the common terms using substitution and try to get the value of (3cos⁡β−5sin⁡β)2{{\left( 3\cos \beta -5\sin \beta \right)}^{2}}.

Complete step by step answer:
The equation given to us is 3sin⁡β+5cos⁡β=53\sin \beta +5\cos \beta =5.
We will apply squares on both sides of the equation.
⇒(3sin⁡β+5cos⁡β)2=52\Rightarrow {{\left( 3\sin \beta +5\cos \beta \right)}^{2}}={{5}^{2}}
We will expand the left hand side using the identity (a+b)2=a2+b2+2ac{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ac.
Therefore, the equation modifies as 9sin⁡2β+25cos⁡2β+2(3)(5)sin⁡βcos⁡β=259{{\sin }^{2}}\beta +25{{\cos }^{2}}\beta +2\left( 3 \right)\left( 5 \right)\sin \beta \cos \beta =25
We will rearrange the equation as follows:
2(3)(5)sin⁡βcos⁡β=25−[9sin⁡2β+25cos⁡2β]......(1)2\left( 3 \right)\left( 5 \right)\sin \beta \cos \beta =25-\left[ 9{{\sin }^{2}}\beta +25{{\cos }^{2}}\beta \right]......\left( 1 \right)
It is given that we have to find the value of (3cos⁡β−5sin⁡β)2{{\left( 3\cos \beta -5\sin \beta \right)}^{2}}.
First of all, we will expand this expression using the identity (a−b)2=a2+b2−2ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab.
(3cos⁡β−5sin⁡β)2=9cos⁡2β+25sin⁡2β−2(3)(5)sin⁡βcos⁡β......(2){{\left( 3\cos \beta -5\sin \beta \right)}^{2}}=9\cos^2 \beta +25{{\sin }^{2}}\beta -2\left( 3 \right)\left( 5 \right)\sin \beta \cos \beta ......\left( 2 \right)
But from (1), we know that 2(3)(5)sin⁡βcos⁡β=25−[9sin⁡2β+25cos⁡2β]2\left( 3 \right)\left( 5 \right)\sin \beta \cos \beta =25-\left[ 9{{\sin }^{2}}\beta +25{{\cos }^{2}}\beta \right].
Therefore, we will make the substitution from (1) into the equation (2).
The equation will modify as follows:
(3cos⁡β−5sin⁡β)2=9cos⁡2β+25sin⁡2β−(25−[(9sin⁡2β+25cos⁡2β)]){{\left( 3\cos \beta -5\sin \beta \right)}^{2}}=9\cos^2 \beta +25{{\sin }^{2}}\beta -\left( 25-\left[\left( 9{{\sin }^{2}}\beta +25{{\cos }^{2}}\beta \right) \right] \right)
We will solve the first parenthesis by multiplying the negative sign.
(3cos⁡β−5sin⁡β)2=9cos⁡2β+25sin⁡2β−25+[9sin⁡2β+25cos⁡2β]{{\left( 3\cos \beta -5\sin \beta \right)}^{2}}=9\cos^2 \beta +25{{\sin }^{2}}\beta -25+\left[ 9{{\sin }^{2}}\beta +25{{\cos }^{2}}\beta \right]
Therefore, the equation changes that (3cos⁡β−5sin⁡β)2=9cos⁡2β+25sin⁡2β−25+9sin⁡2β+25cos⁡2β{{\left( 3\cos \beta -5\sin \beta \right)}^{2}}=9\cos^2 \beta +25{{\sin }^{2}}\beta -25+9{{\sin }^{2}}\beta +25{{\cos }^{2}}\beta
We will rearrange the right hand side so that terms with similar coefficient are together.
(3cos⁡β−5sin⁡β)2=9cos⁡2β+25cos⁡2β+25sin⁡2β+9sin⁡2β−25{{\left( 3\cos \beta -5\sin \beta \right)}^{2}}=9{{\cos }^{2}}\beta +25{{\cos }^{2}}\beta +25{{\sin }^{2}}\beta +9{{\sin }^{2}}\beta -25
Now, we shall take the variables as common on the right hand side.
(3cos⁡β−5sin⁡β)2=(9+25)cos⁡2β+(25+9)sin⁡2β−25{{\left( 3\cos \beta -5\sin \beta \right)}^{2}}=\left( 9+25 \right){{\cos }^{2}}\beta +\left( 25+9 \right){{\sin }^{2}}\beta -25
We know that 9 + 25 = 34.
Therefore, the equation modifies as:
(3cos⁡β−5sin⁡β)2=34cos⁡2β+34sin⁡2β−25{{\left( 3\cos \beta -5\sin \beta \right)}^{2}}=34{{\cos }^{2}}\beta +34{{\sin }^{2}}\beta -25
We will take 34 common.
(3cos⁡β−5sin⁡β)2=34(cos⁡2β+sin⁡2β)−25{{\left( 3\cos \beta -5\sin \beta \right)}^{2}}=34\left( {{\cos }^{2}}\beta +{{\sin }^{2}}\beta \right)-25
From the trigonometric identities, we know cos⁡2β+sin⁡2β=1{{\cos }^{2}}\beta +{{\sin }^{2}}\beta =1
(3cos⁡β−5sin⁡β)2{{\left( 3\cos \beta -5\sin \beta \right)}^{2}} = 34 – 25
(3cos⁡β−5sin⁡β)2{{\left( 3\cos \beta -5\sin \beta \right)}^{2}} = 9

So, the correct answer is “Option A”.

Note: This is a fairly simple problem which requires basic concepts of trigonometric and polynomials. Students are requested to be careful while rearranging as mistakes can be made while the positive or negative signs while rearrangement.