Question

Prove the following trigonometric identities. If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ − 3 cos θ = ± 3.

Answer 1

Answer with Step-by-step explanation:

Given :  

(3 sinθ + 5cosθ)² =  5²

Squaring on both sides.

(3sinθ)² + (5cosθ)² + 2× 3sinθ 5cosθ = 25

[a + b= a² + b² + 2ab]

9sin²θ + 25cos²θ + 30sinθcosθ = 25

9 (1 - cos²θ) + 25(1- sin²θ) + 30 sinθcosθ = 25

[By using  an identity, (1- cos²θ) = sin²θ ,  (1- sin²θ) = cos²θ]

9 - 9cos²θ + 25 -  25sin²θ + 30 sinθcosθ = 25

9 + 25 - (9cos²θ + 25sin²θ - 30sinθcosθ) = 25

34 - (9cos²θ + 25sin²θ - 30sinθcosθ) = 25

- (25sin²θ + 9cos²θ - 30sinθcosθ) = 25 - 34

(25sin²θ + 9cos²θ - 30 sinθcosθ) = 9

(5sinθ - 3cosθ)² = 9

[By using identity , a² - 2ab + b² = (a -  b)² ]

(5sinθ - 3cosθ) = √9

(5sinθ -  3cosθ) = ±3

L.H.S = R.H.S

HOPE THIS ANSWER WILL HELP YOU…

 

Answer 2

$\boxed{\bold{\mathsf{SOLUTION}}}$

$3sin \theta + 5cos \theta = 5 \\ \\ = > (3sin \theta + 5cos \theta) {}^{2} = {5}^{2} \\ \\ = > 9 {sin}^{2} \theta + 25 {cos}^{2} \theta + 30sin \theta.cos \theta = 25 \\ \\ = > 9(1 - {cos}^{2} \theta) + 25(1 - {sin}^{2} \theta) + 30sin \theta.cos \theta = 25 \\ \\ = > 9 - 9 {cos}^{2} \theta + 25 - 25 {sin}^{2} \theta + 30sin \theta.cos \theta = 25 \\ \\ = > - 9 {cos}^{2} \theta - 25 {sin}^{2} \theta + 30sin \theta.cos \theta = 25 - 34 \\ \\ = > 9 {cos}^{2} \theta + 25 {sin}^{2} \theta - 30sin \theta.cos \theta = 9 \\ \\ = > 25 {sin}^{2} \theta + 9 {cos}^{2} \theta - 30sin \theta.cos \theta = 9 \\ \\ = > (5sin \theta - 3cos \theta) {}^{2} = 9 \\ \\ = > 5sin \theta - 3cos \theta = ±3$

hope it helps☺️