Trigo Ques (repost cuz no answer :C)

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

thanks to those who took their time C:

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 ?

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 :)

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.        I still want you to see mine though  

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

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 :)

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

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:

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