Question

If (2sin theta + 3cos theta) = 2, find value of :
(3sin theta - 2cos theta)
Ch - Introduction To Trigonometry
Class 10th ​

Answer 1

Answer :-

3, -3 .

Given:-

$\: \: \: \: \bf \: 2sin \theta + 3cos \theta = 2$

To find :-

$\: \: \: \: \: \bf \: 3sin \theta - 2cos \theta$

Solution :-

Let's consider the value of 3sinθ-2cosθ= k

and,

We have the equation, 2sinθ+3cosθ=2 we shall do squaring on both sides.

$\: \dashrightarrow \bf \:( 2sin \theta + 3cos \theta) {}^{2} =( 2) {}^{2}$

Expanding through (a+b)² = a² + 2ab + b²

$\: \dashrightarrow \bf \: 4sin {}^{2} \theta + 9cos {}^{2} \theta + 12sin \theta \: cos \theta =4$

$\: \dashrightarrow \bf \: \red{ 12sin \theta \: cos \theta = 4 - 4sin {}^{2} \theta - 9cos {}^{2} \theta}$

Now,

$\: \: \: \: \: \bf \: 3sin \theta - 2cos \theta = k$

We shall do squaring on both sides for this equation also

$\dashrightarrow( \bf \: 3sin \theta - 2cos \theta ) {}^{2} =( k) {}^{2}$

$\dashrightarrow \bf \: 9sin {}^{2} \theta + 4cos {}^{2} \theta - 12sin \theta \: cos \theta = k{}^{2}$

We know ,

12sinθcosθ = 4-4sin²θ-9cos²θ

Substituting it

$\dashrightarrow \bf \: 9sin {}^{2} \theta + 4cos {}^{2} \theta - (4 - 4sin {}^{2} \theta - 9cos {}^{2} \theta = k{}^{2}$

$\dashrightarrow \bf \: 9sin {}^{2} \theta + 4cos {}^{2} \theta - 4 + 4sin {}^{2} \theta + 9cos {}^{2} \theta = k{}^{2}$

Regrouping same terms together.

$\dashrightarrow \bf \: 9sin {}^{2} \theta +4sin {}^{2} \theta + 4cos {}^{2} \theta + 9cos {}^{2} \theta - 4 = k{}^{2}$

$\dashrightarrow \bf \: 13sin {}^{2} \theta + 13cos {}^{2} \theta - 4 = k{}^{2}$

$\dashrightarrow \bf \: 13(sin {}^{2} \theta + cos {}^{2} \theta ) - 4 = k{}^{2}$

$\dashrightarrow \bf \: 13( \red {1} ) - 4 = k{}^{2}$

$\dashrightarrow \bf \: 9= k{}^{2}$

$\dashrightarrow \bf \: \pm \sqrt{9} = k{}$

$\dashrightarrow \bf \: k{} = \pm3$

We know that, 3sinθ-2cosθ= k

So,

3sinθ-2cosθ = 3, -3

Answer 2

❐ EXPLANATION:

.

$2 \sin(\theta ) + 3 \cos(\theta ) = 2 \: \: \: ...(given)$

$\sf [Squaring \: Both \: Sides]$

$\implies {(2 \sin(\theta ) + 3 \cos(\theta ) )}^{2} = {2}^{2}$

$\sf [(x + y) ^{2} = {x}^{2} + 2xy + {y}^{2} ]$

$\implies {(2 \sin(\theta ))^{2} + 2(2 \sin(\theta )).(3 \cos(\theta )) + (3 \cos(\theta ) )}^{2} = 4$

$\implies 4 \sin^{2} (\theta ) + 12 \sin(\theta ). \cos(\theta ) + 9 \cos ^{2} (\theta ) = 4$

$\implies 4 \sin^{2} (\theta ) + 4\cos ^{2} (\theta ) + + 5\cos ^{2} (\theta ) + 12 \sin(\theta ). \cos(\theta ) = 4$

$\implies 4( \sin^{2} (\theta ) + \cos ^{2} (\theta ) )+ 5\cos ^{2} (\theta ) + 12 \sin(\theta ). \cos(\theta ) = 4$

$[\sin^{2} (\theta ) + \cos ^{2} (\theta )= 1 ]$

$\implies 4( 1)+ 5\cos ^{2} (\theta ) + 12 \sin(\theta ). \cos(\theta ) = 4$

$\implies \cancel{4} + 5\cos ^{2} (\theta ) + 12 \sin(\theta ). \cos(\theta ) = \cancel{4}$

$\implies 5\cos ^{2} (\theta ) = - 12 \sin(\theta ). \cos(\theta ) \: \: \:... (1)$

$3 \sin(\theta ) - 2 \cos(\theta) = 3 \sin(\theta ) - 2 \cos(\theta)$

$\sf [Squaring \: Both \: Sides]$

$\implies {(3 \sin(\theta ) - 2 \cos(\theta) )}^{2} = {(3 \sin(\theta ) - 2 \cos(\theta) )}^{2}$

$\sf [(x - y) ^{2} = {x}^{2} - 2xy + {y}^{2} ]$

$\implies {(3 \sin(\theta ) - 2 \cos(\theta) )}^{2} = {(3 \sin(\theta ))}^{2} - 2(3 \sin(\theta ))(2 \cos(\theta)) + {(2 \cos(\theta) )}^{2}$

$\implies {(3 \sin(\theta ) - 2 \cos(\theta) )}^{2} = {9\sin^{2} (\theta )} - 12 \sin(\theta ). \cos(\theta) + {4\cos^{2} (\theta)}$

$\implies {(3 \sin(\theta ) - 2 \cos(\theta) )}^{2} = {9\sin^{2} (\theta )} + 5\cos^{2} (\theta) + {4\cos^{2} (\theta)} \: \: \: ...(By \: Eq \: [1])$

$\implies {(3 \sin(\theta ) - 2 \cos(\theta) )}^{2} = 9\sin^{2} (\theta ) + 9\cos^{2} (\theta)$

$\implies {(3 \sin(\theta ) - 2 \cos(\theta) )}^{2} = 9(\sin^{2} (\theta ) + \cos^{2} (\theta) )$

$[\sin^{2} (\theta ) + \cos ^{2} (\theta )= 1 ]$

$\implies {(3 \sin(\theta ) - 2 \cos(\theta) )}^{2} = 9(1 )$

$\implies {3 \sin(\theta ) - 2 \cos(\theta) } = \sqrt{9}$

$\implies \red{\boxed {{3 \sin(\theta ) - 2 \cos(\theta) } = ± 3 }}$