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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?

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Kevin S. |
5.0 5.0 (4 lesson ratings) (4)
2

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...

 

 

Comments

Nice, 

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

Matthew S. | Statistics, Algebra, Math, Computer Programming TutorStatistics, Algebra, Math, Computer Prog...
4.9 4.9 (22 lesson ratings) (22)
1

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.

Sam M. | Mr. Gets ResultsMr. Gets Results
4.8 4.8 (119 lesson ratings) (119)
0

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