Question

3 sin theta+4cos theta = 5,prove that sin theta=3/5​

Answer 1

Given ---->

$3 \: sin \alpha \: + \: 4 \: cos \alpha \: = 5$

To prove ---->

$sin \alpha \: = \dfrac{3}{5}$

Concept used---->

1)

$\blue{{\large( \: a \: - \: b \: ) ^{2} = \: {a}^{2} \: + \: {b}^{2} \: - 2 \: a \: b}}$

2)

$\green{{\large1 \: - \: {sin}^{2} \alpha \: = {cos}^{2} \alpha}}$

Solution-----> ATQ,

$3 \: sin \alpha \: + \: 4 \: cos \alpha \: = \: 5$

=>$4 \: cos \alpha \: = \: 5 \: - \: 3 \: sin \alpha$

$squaring \: both \: sides \: we \: get$

=>$( \: 4 \: cos \alpha \: ) ^{2} \: = ( \: 5 \: - \: 3 \: sin \alpha \: ) ^{2}$

$= > 16 \: {cos}^{2} \alpha \: = ( \: 5 \: ) ^{2} \: + ( \: 3 \: sin \alpha ) ^{2} \: - \: 2 \: ( \: 5 \: ) \: ( \: 3sin \alpha )$

$= > 16 \: ( \: 1 \: - {sin}^{2} \alpha \: ) \: = 25 \: + 9 {sin}^{2} \alpha \: - 30 \: sin \alpha$

$= > 16 \: - \: 16 {sin}^{2} \alpha \: = 25 \: + 9 {sin}^{2} \alpha \: - 30 \: sin \alpha$

$= > 25 \: {sin}^{2} \alpha - \: 30 \:sin \alpha \: + 9 \: = 0$

$= > \: ( \: 5 \: sin \alpha \: ) ^{2} \: - 2 \: ( \: 5 \: sin \alpha \: ) \: ( \: 3 \: ) \: + \: ( \: 3 \: ) ^{2} \: = 0$

$= > \: ( \: 5 \: sin \alpha \: - \: 3 \:) ^{2} \: = \: 0$

$taking \: square \: root \: both \: sides \:$

$= > \: 5 \: sin \alpha \: - \: 3 \: = 0$

$= > 5 \: sin \alpha \: = \: 3$

$= > \: sin \alpha \: = \: \dfrac{3}{5}$

Answer 2

Answer:

3/5

Step-by-step explanation:

sin theta= 3/5 at first do the whole square and then make it a perfect square so that at the 2nd last line (5 sin theta-3)^2 = 0 come by which it become easier to solve .