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Given that x + sin y = 2008 and x + 2008 cos y = 2007 where 0 ≤ y ≤ π/2. The value of [x + y], is. (Here [x] denotes greatest integer function)
- a)2007
- b)2008
- c)2009
- d)2010
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
Given that x + sin y = 2008 and x + 2008 cos y = 2007 where 0 ≤ y &...
x + sin y = 2008
This is possible only if cos y = 0
This is possible only if cos y = 0
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Given that x + sin y = 2008 and x + 2008 cos y = 2007 where 0 ≤ y &...
We can solve this system of equations by dividing the two equations:
\[\frac{x \sin y}{x \cos y} = \frac{2008}{2007}\]
Simplifying, we have:
\[\tan y = \frac{2008}{2007}\]
Taking the inverse tangent of both sides, we find:
\[y = \tan^{-1} \left(\frac{2008}{2007}\right)\]
Now, we can substitute this value of $y$ into one of the original equations to solve for $x$. Let's use the first equation:
\[x \sin \left(\tan^{-1} \left(\frac{2008}{2007}\right)\right) = 2008\]
Since $\sin$ and $\tan^{-1}$ are inverse functions, we have:
\[x \cdot \frac{2008}{\sqrt{2008^2 + 2007^2}} = 2008\]
Simplifying, we find:
\[x = \frac{2008 \cdot \sqrt{2008^2 + 2007^2}}{2008} = \sqrt{2008^2 + 2007^2}\]
Therefore, $x = \sqrt{2008^2 + 2007^2}$.
\[\frac{x \sin y}{x \cos y} = \frac{2008}{2007}\]
Simplifying, we have:
\[\tan y = \frac{2008}{2007}\]
Taking the inverse tangent of both sides, we find:
\[y = \tan^{-1} \left(\frac{2008}{2007}\right)\]
Now, we can substitute this value of $y$ into one of the original equations to solve for $x$. Let's use the first equation:
\[x \sin \left(\tan^{-1} \left(\frac{2008}{2007}\right)\right) = 2008\]
Since $\sin$ and $\tan^{-1}$ are inverse functions, we have:
\[x \cdot \frac{2008}{\sqrt{2008^2 + 2007^2}} = 2008\]
Simplifying, we find:
\[x = \frac{2008 \cdot \sqrt{2008^2 + 2007^2}}{2008} = \sqrt{2008^2 + 2007^2}\]
Therefore, $x = \sqrt{2008^2 + 2007^2}$.
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