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If $3sin\theta + 5cos\theta = 5$, prove that $5sin\theta - 3cos\theta = \pm 3.$

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Hint- Let ${\text{5}}sin\theta - 3cos\theta = x$ then square both equations and add them .

We have been given that
$3sin\theta + 5cos\theta = 5$
Now let ${\text{5}}sin\theta - 3cos\theta = x$
When we square both the equations, we get
$9si{n^2}\theta + 25co{s^2}\theta + 30sin\theta cos\theta = 25$ - Equation (1)
And
$25si{n^2}\theta + 9co{s^2}\theta - 30sin\theta cos\theta = {x^2}$ - Equation (2)
Now if we add Equation (1) and Equation (2), we get
$
  34(si{n^2}\theta + co{s^2}\theta ) = 25 + {x^2} \\
   \Rightarrow {{\text{x}}^2} = 34 - 25{\text{ }}\left( {{\text{As }}si{n^2}\theta + co{s^2}\theta = 1} \right) \\
   \Rightarrow {{\text{x}}^2} = 9 \\
   \Rightarrow x = \pm 3 \\
$
This gives us that
$5sin\theta - 3cos\theta = \pm 3$

Note- In these types of questions the catch is that we must square and add the equations to solve them.
 This method solves the equations in the fastest and easiest way, then using a simple trigonometric
formula we can easily prove the required expression.
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