+0

Trigo Ques (repost cuz no answer :C)

+1
142
7

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
Sort:

7+0 Answers

#1
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
+89752
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
+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
+89752
+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
+89752
+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
+18352
+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

5 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details