Question

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Answer 1

Question :

If 3sinθ+4cosθ= 5

so prove that

sinθ=$\dfrac{3}{5}$

Formula's Used :

1) sin(A+B)=sinAcosB+sinBcosA

2) sin(A-B) =sinAcosB-sinBcosA

3) cos(A-B)=cosAcosB+sinAsinB

4)cos(A+B)=cosAcosB-sinAsinB

Solution :

Given : 3sinθ+4cosθ= 5

we have to Prove that sinθ=$\dfrac{3}{5}$

_____________________________

3sinθ+4cosθ= 5

Divide both sides by 5

$\implies \frac{3}{5} \sin( \theta) + \frac{4}{5} \cos( \theta) = 1$

Now let sinA=$\dfrac{3}{5}$

⇒cosA=√1-sin²A= $\dfrac{4 }{5}$

$\implies \: \sin(A) \sin( \theta) + \cos(A) \cos( \theta) = 1$

$\implies \: \cos(A - \theta) = 1$

we know that cos 0° = 1

$\implies \: A- \theta = 0$

$\implies\theta = A$

Now take sin on both sides

$\implies \sin( \theta) = \sin(A)$

$\implies \: \sin( \theta) = \frac{3}{5}$

$\huge{\bold{ Hence\: Proved}}$