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Heya

Here's the ques:

 

Given : sin\alpha+sin\beta=1 , cos\alpha+cos\beta=1

Find the value of sin\alpha-cos\beta

 

Thanks~ 

 

 

Thanks for taking your time....

Guest May 13, 2017
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7+0 Answers

 #1
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0

Nevermind, I solved it. took 8 hrs, but I solved it

 

thanks to those who took their time C:

Guest May 14, 2017
 #2
avatar+90526 
0

Given : sin\alpha+sin\beta=1 , cos\alpha+cos\beta=1

Find the value of sin\alpha-cos\beta

 

\(Given : \\sin\alpha+sin\beta=1 , \\cos\alpha+cos\beta=1\\ \text{Find the value of }sin\alpha-cos\beta\\ \\~\\ \)

 

I've been playing with it without success.

Could you please give an outline of how you solved it ?

Melody  May 14, 2017
edited by Melody  May 14, 2017
 #3
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+2

Sure Melody~ C:

 

2 methods (because I'm not sure if the first method is viable):

 

1st method:

square both equations and add them together   \(2sin\alpha sin\beta+2cos\alpha cos\beta=0\)

which is \(cos(\alpha-\beta)=0\)

\(\alpha-\beta=arccos0 \)

\(\alpha-\beta=\pi/2\)

\(\alpha=\beta+\pi/2\)

substitute this into the first equation and subtract it from the first equation and you can get \(sin\alpha-cos\beta=0\)

The reason why I'm not sure whether this method is correct or not is because there are more angles of theta in which cos theta =0

 

So I came up with this 2nd method:

convert both sin beta and cos alpha into \(\sqrt{1-{cos}^{2}\beta} and \sqrt{1-{sin}^{2}\alpha}\)respectively

move all the non-square roots to one side and square them

You will get two equations that have the same LHS/RHS, which are

\({sin}^{2}\alpha+{cos}^{2}\beta=2cos\beta\)

\({sin}^{2}\alpha+{cos}^{2}\beta=2sin\alpha\)

And you probably know the rest!

 

P.S. This is a past-year question from my school's trigonometry exam.....oh and can you tag users in the forum? That would be helpful

Guest May 14, 2017
 #4
avatar+90526 
+1

Thanks very much for that :)

 

I also came up with your first answer but I was not happy with it either.

 

I haven't not looked properly at your second answer yet.  But I will :)

 

I am not sure what you mean by tag people but if you are a member there are some things you can do.

If you get into a persons member page you can view all their questions and answers. I expecxt you could bookmark this page although I have never tried to do that.

If you are going to use this site more than once or twice it really is worth your while to become a member :)

Cheers :)

Melody  May 14, 2017
 #5
avatar+90526 
+1

 

It seems to me that before I tacklyethis I need to work out what the restrictions are on alpha and beta.

I am just going to elaborate on your first method first :)

I wish you (the question asker and answerer) were a member because then I would be much more sure that you would at least see I have done another answer :(    

Yes, I know you already have the answer. frown       I still want you to see mine though  laugh

 

\(Given : \\sin\alpha+sin\beta=1 , \\cos\alpha+cos\beta=1\\ \text{Find the value of }sin\alpha-cos\beta\\ \\~\\ \text{since ALL sin and cos values must be between -1 and +1 AND} \\sin\alpha+sin\beta=1 \;\;\\\text{It follows that }sin\alpha \;\;and\;\; sin\beta \text { must both be between 0 and 1 }\\ so\;\; \alpha \;and\; \beta \;\text{must both be in the first or second quadrant}\\ cos\alpha+cos\beta=1 \\\text{It follows that }cos\alpha \;\;and\;\; cos\beta \text { must both be between 0 and 1 }\\ so\;\; \alpha \;and\; \beta \;\text{must both in the first quadrant}\\ \text{HENCE: }sin\alpha, \;\;sin\beta\; \;cos\alpha,\; cos\beta\; \text {are all between 0 and 1} \)

 

Squaring  both original equations, which we can do becasue everything is positive we have:

 

\(sin^2\alpha+sin^2\beta+2sin\alpha sin\beta=1\\ cos^2\alpha+cos^2\beta+2cos\alpha\cos\beta=1\\ add\\ 1+1+2(sin\alpha sin\beta+cos\alpha cos\beta)=2\\ 2(sin\alpha sin\beta+cos\alpha cos\beta)=0\\ sin\alpha sin\beta+cos\alpha cos\beta=0\\ cos(\alpha-\beta)=cos(\beta-\alpha)=0 \quad Let\;\beta\ge \alpha \quad (I\;mean\;the\; 0 \;\;to\;\; \frac{\pi}{2}\; \text {equivalent angle)}\\ cos(\beta-\alpha)=0\\ \color{blue}{\text{From here on I am going to use n as an integer but its value can be different each time}}\\ \beta-\alpha=\frac{\pi}{2}+2\pi n \qquad n\in Z\\ \beta=\alpha+\frac{\pi}{2}+2\pi n \qquad n\in Z\\ \text{But }\beta \text{ must be in the first quadrant so }\alpha = 0+2\pi n \;\;and\;\;\beta=\frac{\pi}{2}+2\pi n \\ so\\ sin\alpha - \cos\beta=sin(2\pi n)- cos(\frac{\pi}{2}+2\pi n)\\ sin\alpha - \cos\beta=0- 0\\ sin\alpha - \cos\beta=0\\ \)

I think that method will stand up to scrutiny.   I haven't evewn looked at your second answer yet :)

Melody  May 14, 2017
 #6
avatar
+1

Here's another method.

\(\displaystyle \sin\alpha + \sin\beta=2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)=1\).

\(\displaystyle \cos\alpha+\cos\beta=2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)=1\).

Divide the first by the second,

\(\displaystyle \tan\left(\frac{\alpha+\beta}{2}\right)=1\),

and therefore, assuming that the angles are acute,

\(\displaystyle \alpha+\beta=90\deg\),

in which case

\(\displaystyle \sin\alpha = \cos\beta \qquad \text{so}\qquad \sin\alpha-\cos\beta=0\).

 

The equation for the tangent has multiple solutions,

\(\displaystyle \frac{\alpha+\beta}{2}= 180n+45 \deg \quad \text{n an integer}\).

All lead to the same conclusion.

 

Tiggsy

Guest May 14, 2017
 #7
avatar+18605 
+1

Given:

\(sin\alpha+sin\beta=1,\ cos\alpha+cos\beta=1\)

Find the value of

\(sin\alpha-cos\beta\)

 

\(\begin{array}{|lrcll|} \hline (1) & \sin(\alpha) + \sin(\beta) &=& 1 \\ (2) & \cos(\alpha) + \cos(\beta) &=& 1 \\ \hline (1)-(2): & \sin(\alpha) + \sin(\beta) - \Big(\cos(\alpha) + \cos(\beta) \Big) &=& 0 \\ & \sin(\alpha) + \sin(\beta) -\cos(\alpha) - \cos(\beta) &=& 0 \\ & \underbrace{\sin(\alpha) -\cos(\alpha)}_{=\sqrt{2}\sin(\alpha-45^{\circ})} &=& \underbrace{\cos(\beta) - \sin(\beta)}_{=-\sqrt{2}\sin(\beta-45^{\circ})} \\ & \sqrt{2}\sin(\alpha-45^{\circ}) &=& -\sqrt{2}\sin(\beta-45^{\circ}) \\ & \sin(\alpha-45^{\circ}) &=& - \sin(\beta-45^{\circ}) \\ & \sin(\alpha-45^{\circ}) &=& \sin\Big(-(\beta-45^{\circ})\Big) \\ & \sin(\alpha-45^{\circ}) &=& \sin(45^{\circ}-\beta ) \\ & \alpha-45^{\circ} &=& 45^{\circ}-\beta \\ & \alpha &=& 90^{\circ}-\beta \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && \sin(\alpha) - \cos(\beta) \\ & =& \sin(90^{\circ}-\beta) - \cos(\beta) \\ & =& \cos(\beta) - \cos(\beta) \\ & =& 0 \\ \hline \end{array}\)

 

Proof:

\(\begin{array}{|rcll|} \hline \sin(\alpha) + \sin(\beta) &=& 1 \quad | \quad \alpha = 90^{\circ}-\beta \\ \sin(90^{\circ}-\beta) + \sin(\beta) &=& 1 \\ \cos(\beta) + \sin(\beta) &=& 1 \quad | \quad \cos(\beta) + \sin(\beta)=\sqrt{2}\sin(\beta+45^{\circ}) \\ \sqrt{2}\sin(\beta+45^{\circ}) &=& 1 \\ \sin(\beta+45^{\circ}) &=& \frac{1}{\sqrt{2}} \\ \beta+45^{\circ} &=& \arcsin( \frac{1}{\sqrt{2}} ) + n\cdot 360^{\circ} \qquad n \in \mathbb{Z} \\ \beta+45^{\circ} &=& 45^{\circ} + n\cdot 360^{\circ} \qquad n \in \mathbb{Z} \\ \beta &=& 90^{\circ} + n\cdot 360^{\circ} \qquad n \in \mathbb{Z} \\\\ \alpha &=& 90^{\circ}-\beta \\ \alpha &=& 90^{\circ}-(90^{\circ} + n\cdot 360^{\circ}) \qquad n \in \mathbb{Z} \\ \alpha &=& 0^{\circ} + n\cdot 360^{\circ} \qquad n \in \mathbb{Z} \\ \hline \end{array}\)

 

 

laugh

heureka  May 15, 2017

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