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Question: If \[3\sin \theta + 5\cos \theta = 5\], then \[5\sin \theta - 3\cos \theta \] is equal to A. \[4\...

If 3sin⁡θ+5cos⁡θ=53\sin \theta + 5\cos \theta = 5, then 5sin⁡θ−3cos⁡θ5\sin \theta - 3\cos \theta is equal to
A. 44
B. ±3 \pm 3
C. 55
D. None of the above

Explanation

Solution

We can solve this using simple algebraic identities. We square on both sides of 3sin⁡θ+5cos⁡θ=53\sin \theta + 5\cos \theta = 5. We expand it using algebraic identity (a+b)2=a2+b2+2ab{\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⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 we will have the desired result.

Complete step by step answer:
Given
3sin⁡θ+5cos⁡θ=53\sin \theta + 5\cos \theta = 5
Now squaring on both sides we have,
(3sin⁡θ+5cos⁡θ)2=52{\left( {3\sin \theta + 5\cos \theta } \right)^2} = {5^2}
Now applying the algebraic identity (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab, we have
(3sin⁡θ)2+(5cos⁡θ)2+2(3sin⁡θ)(5cos⁡θ)=52{\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}
9sin⁡2θ+25cos⁡2θ+30sin⁡θcos⁡θ=259{\sin ^2}\theta + 25{\cos ^2}\theta + 30\sin \theta \cos \theta = 25
Now from Pythagoras identity we have sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1, using this we have sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta and cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta . Applying this we have
9(1−cos⁡2θ)+25(1−sin⁡2θ)+30sin⁡θcos⁡θ=259\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
9−9cos⁡2θ+25−25sin⁡2θ+30sin⁡θcos⁡θ=259 - 9{\cos ^2}\theta + 25 - 25{\sin ^2}\theta + 30\sin \theta \cos \theta = 25
Now grouping we have
9+25−9cos⁡2θ−25sin⁡2θ+30sin⁡θcos⁡θ=259 + 25 - 9{\cos ^2}\theta - 25{\sin ^2}\theta + 30\sin \theta \cos \theta = 25
Or
9+25−(9cos⁡2θ+25sin⁡2θ−30sin⁡θcos⁡θ)=259 + 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−(9cos⁡2θ+25sin⁡2θ−30sin⁡θcos⁡θ)=09 - \left( {9{{\cos }^2}\theta + 25{{\sin }^2}\theta - 30\sin \theta \cos \theta } \right) = 0
Shifting the terms we have
(9cos⁡2θ+25sin⁡2θ−30sin⁡θcos⁡θ)=9\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 (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab, where a=3cos⁡θa = 3\cos \theta and b=5sin⁡θb = 5\sin \theta . Applying this we have
(3cos⁡θ−5sin⁡θ)2=9{\left( {3\cos \theta - 5\sin \theta } \right)^2} = 9
Now taking square root on both sides we have
(3cos⁡θ−5sin⁡θ)=±9\left( {3\cos \theta - 5\sin \theta } \right) = \pm \sqrt 9
We know that the sure root of 9 is 3. Then
⇒3cos⁡θ−5sin⁡θ=±3\Rightarrow 3\cos \theta - 5\sin \theta = \pm 3.
⇒5sin⁡θ−3cos⁡θ=±3\Rightarrow 5\sin \theta - 3\cos \theta = \pm 3.
Thus the required answer is ±3 \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⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1, tan⁡2θ+1=sec⁡2θ{\tan ^2}\theta + 1 = {\sec ^2}\theta and 1+cot⁡2θ=csc⁡2θ1 + {\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.