Question

If $3sin\theta+5cos\theta=5$ then show that $5sin\theta-3cos\theta=\pm3$​

Answer 1

=>( 3 sin θ + 5cos θ )² + ( 5 sinθ - 3 cos θ )² .

= 9( sin²θ + cos²θ ) + 25( sin²θ + cos²θ ) .

= ( 9 + 25 ) .

= 34 .

∴ ( 3 sin θ + 5 sin θ )² + ( 5 sin θ - 3 cos θ )² = 34 .

⇒ 5² + ( 5 sin θ - 3 cos θ )² = 34 . [ ∵ 3 sin θ + 5 cos θ = 5 ]

⇒ 25 + ( 5 sin θ - 3 cos θ )² = 34 .

⇒ ( 5 sin θ - 3 cos θ )² = 34 - 25 .

⇒ ( 5 sin θ - 3 cos θ )² = 9 .

⇒ ( 5 sin θ - 3 cos θ ) = ±√9 .

⇒ ( 5 sin θ - 3 cos θ ) = ± 3 .

Hence, ( 5 sin θ - 3 cos θ ) = ± 3 .

Answer 2

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$3 \sin\theta + 5cos \theta = 5 \\ \\ \\ {(3sin \theta + 5cos \theta)}^{2} = 25 \\ \\ \\9 {sin}^{2} \theta + 25 {cos}^{2} \theta \: + 15sin \theta cos \theta = 25 \\ \\ \\9( {sin}^{2} \theta + {cos}^{2} \theta) + 16 {cos}^{2} \theta + 15sin \theta cos \theta = 25 \\ \\ \\9 + 16 {cos}^{2} \theta + 15sin \theta cos \theta = 25 \\ \\ \\16 {cos}^{2} \theta + 15sin \theta cos \theta =25 - 9 = 16 \\ \\ \\15sin \theta cos \theta = 16 - 16 {cos}^{2} \theta \\ \\ \\15sin \theta cos \theta = 16(1 - {cos}^{2} \theta \\ \\ \\15sin \theta cos \theta = 16 {sin}^{2} \theta$

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${(5sin \theta - 3cos \theta)}^{2} \\ \\ = 25 {sin}^{2} \theta + 9 {cos}^{2} \theta -15sin \theta cos \theta \\ \\ = 9( {sin}^{2} \theta + {cos}^{2} \theta) + 16 {sin}^{2} \theta - 15sin \theta cos \theta \\ \\ \\ = 9 + 16 {sin}^{2} \theta - 16 {sin}^{2} \theta \\ \\ \\ = 9$

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Hence,

$5sin \theta - 3cos \theta = \pm \sqrt{9} \\ \\ \\5sin \theta - 3cos \theta = \pm3$

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Hence, Proved!

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