Question

if 3sin0+4 cos0=5 , find sin0​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: If 3sin\theta \: + 4cos\theta \: = 5, \: find \:sin\theta \:$

$\large\underline{\bold{Solution-}}$

Identities Used :-

$1. \: \: \: \boxed{ \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }}$

$2. \: \: \: \boxed{ \bf{ {sin}^{2}\theta \: + {cos}^{2}\theta \: = 1 }}$

$3. \: \: \: \boxed{ \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }}$

Let solve the problem now!!

Given that

$\sf \: 3sin\theta \: + 4cos\theta \: = 5$

$\sf \:4cos\theta \: = 5 - 3sin\theta \:$

On squaring both sides, we get

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

$\sf \: 16 {cos}^{2} \theta \: = 25 + {9sin}^{2} \theta \: - 30sin\theta \:$

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

$\sf \: 16 - 16 {sin}^{2} \theta \: = 25 + {9sin}^{2} \theta \: - 30sin\theta \:$

$\sf \: 25 {sin}^{2} \theta \: - 30sin\theta \: + 9 = 0$

$\sf \: {(5sin\theta \:)}^{2} + {(3)}^{2} - 2 \times 5sin\theta \: \times 3 = 0$

$\sf \: {(5sin\theta \: - 3)}^{2} = 0$

$\sf \: \: 5sin\theta \: - 3 = 0$

$\: \: \: \: \therefore \boxed{\bf \: sin\theta \: = \dfrac{3}{5} }$

Additional Information :-

$1. \: \: \: \boxed{ \bf{ {sec}^{2}\theta \: - {tan}^{2}\theta \: = 1 }}$

$2. \: \: \: \boxed{ \bf{ {cosec}^{2}\theta \: - {cot}^{2}\theta \: = 1 }}$

$3. \: \: \: \boxed{ \bf{cosec\theta \: - cot\theta \: = \dfrac{1}{cosec\theta \: + cot\theta \:} }}$

$4. \: \: \: \boxed{ \bf{cosec\theta \: + cot \theta \:= \dfrac{1}{cosec\theta \: - cot\theta \:} }}$

$5. \: \: \: \boxed{ \bf{sec\theta \: + tan\theta \: = \dfrac{1}{ sec\theta \: - tan\theta \: } }}$

$6. \: \: \: \boxed{ \bf{sec\theta \: - tan\theta \: = \dfrac{1}{ sec\theta \: + tan\theta \: } }}$