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# Trigo Ques (repost cuz no answer :C)

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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~

Guest May 13, 2017
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#1
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Nevermind, I solved it. took 8 hrs, but I solved it

thanks to those who took their time C:

Guest May 14, 2017
#2
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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
+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
+91972
+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
+91972
+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.        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 :)

Melody  May 14, 2017
#6
+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
+19096
+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:

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

heureka  May 15, 2017

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