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# Trigonometry

Let S be the sum of all the possible values of sinx that satisfy the following equation:

5−2cos2x−7sinx=0
S can be written as ab, where a and b are coprime positive integers. What is the value of a+b?

First, let's get everything in terms of one function. Let's replace cos 2x with the double angle formula
cos 2x = 1 - 2 sin2x

5 - 2 (1 - 2sin2x) - 7 sin x = 0  . Simplifying, we get
5 - 2 + 4 sin2x - 7 sinx = 0, then
3 + 4 sin2x - 7 sinx = 0

Rearranging terms, we get
4 sin2x - 7 sinx + 3 = 0

Hmmm... this looks a lot like a quadratic equation... let's factor it

(4 sin x - 3)(sin x - 1) = 0

sin x - 1 = 0    ===> sin x = 1

4 sin x - 3 = 0 ===>  sin x = 3/4

So if S = the sum of values of sin x, then S would equal 1 3/4, but I'm missing identifying values for a and b that would be positive integers where the product would equal 1 3/4...

I appreciate any input from my esteemed colleagues who can take it from here...

thank u so much..

3/8/2013

Nice,

You're welcome. That's why we're here.

3/16/2013

I'll give it a try...

It looks like the only time this equation can be true is when (cos 2x) evaluates to -1 and, simultaneously, (sin x) evaluates to 1.  In that case 5 - 2(-1) - 7(1) = 0.  Since sine and cosine values range between 1 and -1, this is the only pair that satisfies the equation.  We know (sin x) = 1 at (pi/2) and again at (5*pi)/2, (9*pi)/2, etc.  These are also the points at which cos(2x) = -1; if you graph cos(2x) it is "compressed" such that its periodicity is in pi radians and not (2*pi) radians.

If I'm reading this question right, the only possible value of (sin x) that satisfies this equation is 1.  So S=1, and the only possible way to satisfy ab=1 and have them both be positive integers is for a and b to equal 1.  So a+b=2.

I'm curious to know what class presented you with such a problem!  Unless there are some fancy theories I'm forgetting, or forgot from my college courses 20+ years ago, then please disregard everything I said before.  But this is the only answer I can come up with to satisfy the problem.

Hello.

Kevin S. In Seneca did an excellent job in figuring out that this equation can be factored and set equal to zero.  It can be factored into two binomials:  (4Sin(x) - 3)(Sin(x) - 1) = 0.

From here, it's a piece of cake.  You just set the factors to zero.  Sin(x) - 1 will be equal to zero when Sin(x) is equal to 1.  This occurs at 90 degrees, or pi/2 depending on whether you are in degree or radian measure.  Now as for making 4Sin(x) - 3 equal to zero; first you do some algebra and get that sin(x) = 3/4.  In other words, the value of Sine in this case must be 0.75.  It is important to note, however, that this is NOT A VALUE of Sine which appears explicitly on the unit circle.  It's not, say, the Sine of 2pi/3 or 45 degrees.  So how do you figure it out?  I have three words for you:  GOD BLESS CALCULATORS.

If you type Sin(-1)(.75) into any scientific calculator, you will get your answer.  In radian measure, it's about .848.  In degree measure, it's 48.59 degrees.

Mr. Gets Results