Solve the given trigonometry problem

[chwala]

Homework Statement

See attached

Relevant Equations

Trigonometry

text solution here;

I was solving this today...got stuck and wanted to consult here...but i eventually found the solution...any insight/alternative approach is welcome...
My approach;

...
$\sin^2y+ cos^2 y= 2a^2-2a \sin x - 2a\cos x+1$

It follows that,$2a(\sin x + \cos x)=2a^2$

$\sin x+\cos x=a$

cheers.

 

[anuttarasammyak]

Good job

I would just add a short conjecture. x and y are exchangeable in the formula given and
$$\sin x+\sin y + \cos x + \cos y=2a$$
So we may guess
$$\sin x+\cos x = \sin y + \cos y=a$$

 

[anuttarasammyak]

sin⁡x+cos⁡x=a

By squaring
$$2 \sin x \cos x = a^2-1$$
So$$x,y=\theta, \frac{\pi}{2}-\theta$$where
$$\theta= \frac{1}{2} \sin^{-1}(a^2-1)$$

 

[SammyS]

By squaring $$2 \sin x \cos x = a^2-1$$ So$$x,y=\theta, \frac{\pi}{2}-\theta$$where $$\theta= \frac{1}{2} \sin^{-1}(a^2-1)$$

Your solution is quite interesting.

Of course, the problem stated in the OP only asked that $\sin x +\sin y$ be given in terms of $a$ .

... but to pursue your solution further:

Since $\displaystyle \sin 2x = a^2-1$, the range of values for $a$ is limited if there is to be any real solution for $x$..

 

[brotherbobby]

I solve it in a different way.

I get $\sin x+\cos x = \sqrt 2$. Can someone tell me where am I mistaken?

 

[chwala]

I solve it in a different way.

View attachment 316966I get $\sin x+\cos x = \sqrt 2$. Can someone tell me where am I mistaken?

Mythoughts on this...

We know that

$\sin x + \sin y = 2 \sin \dfrac {x+y}{2} ⋅ \cos \dfrac {x-y}{2}=a$

and

$\cos x + \cos y =2 \cos \dfrac{x+y} {2}⋅\cos \dfrac {x-y}{2}=a$

$⇒4⋅\sin^2 \left[\dfrac{x+y}{2}\right] ⋅ cos^2 \left[\dfrac{x-y}{2}\right]=a^2$

$⇒4⋅\cos^2 \left[\dfrac{x+y}{2}\right] ⋅ cos^2 \left[\dfrac{x-y}{2}\right]=a^2$

on adding the two we end up with;

$cos^2 \left[\dfrac{x-y}{2}\right]\left(4⋅\sin^2 \left[\dfrac{x+y}{2}\right]+4⋅\cos^2 \left[\dfrac{x+y}{2}\right]\right)=2a^2$

$4⋅cos^2 \left[\dfrac{x-y}{2}\right]=2a^2$

$cos^2 \left[\dfrac{x-y}{2}\right]=\dfrac{2a^2}{4}$

$⇒cos\left[\dfrac{x-y}{2}\right]=\dfrac{a}{\sqrt{2}}$

Therefore on substituting back to our equation;

$2 \sin \dfrac {x+y}{2} ⋅ \cos \dfrac {x-y}{2}=a$

and

$2 \cos \dfrac{x+y} {2}⋅\cos \dfrac {x-y}{2}=a$

we get;

$2 \sin \dfrac {x+y}{2} ⋅ \dfrac{a}{\sqrt{2}}=a$

$2 \cos \dfrac{x+y} {2}⋅\dfrac{a}{\sqrt{2}}=a$

$⇒\sin \dfrac {x+y}{2}=\dfrac{\sqrt{2}}{2}$

$⇒\cos \dfrac {x+y}{2}=\dfrac{\sqrt{2}}{2}$

$⇒x+y=\dfrac{π}{2}$

Let $x=y$ then $x=y=\dfrac{π}{4}$

$\sin x +\cos x=\sin \dfrac{π}{4} +\cos \dfrac{π}{4}=\sqrt{2}$

but;
$⇒\cos\left[\dfrac{x-y}{2}\right]=\dfrac{a}{\sqrt{2}}$

since $x=y$ then;

$1=\dfrac{a}{\sqrt{2}}$

$⇒a=\sqrt{2}$

Therefore;

$\sin x + \cos x= a$

 

[anuttarasammyak]

Since sin⁡2x=a2−1, the range of values for a is limited if there is to be any real solution for x..

For real x
$$0 < |a| \leq \sqrt{2}$$ as a $\neq$ 0 in the problem statement.
I wonder whether and why sign of a does not matter with the solution. So as another way without squaring,
$$\sqrt{2}\sin(x+\frac{\pi}{4})=a$$
$$x=\sin^{-1}\frac{a}{\sqrt{2}}-\frac{\pi}{4}$$
with y
$$x,y=\sin^{-1}\frac{a}{\sqrt{2}}-\frac{\pi}{4},-\sin^{-1}\frac{a}{\sqrt{2}}+\frac{3\pi}{4}$$
where sign of a matters.

 

[anuttarasammyak]

I solve it in a different way.

View attachment 316966I get $\sin x+\cos x = \sqrt 2$. Can someone tell me where am I mistaken?

$$x-y=\pi$$
,which gives a=0 the prohibited value, does not have AND relation but OR relation with
$$x+y=\frac{\pi}{2}$$
,which allows a to be variable, in getting the solution.

 

[brotherbobby]

Mythoughts on this...

We know that

$\sin x + \sin y = 2 \sin \dfrac {x+y}{2} ⋅ \cos \dfrac {x-y}{2}=a$

and

$\cos x + \cos y =2 \cos \dfrac{x+y} {2}⋅\cos \dfrac {x-y}{2}=a$

$⇒4⋅\sin^2 \left[\dfrac{x+y}{2}\right] ⋅ cos^2 \left[\dfrac{x-y}{2}\right]=a^2$

$⇒4⋅\cos^2 \left[\dfrac{x+y}{2}\right] ⋅ cos^2 \left[\dfrac{x-y}{2}\right]=a^2$

on adding the two we end up with;

$cos^2 \left[\dfrac{x-y}{2}\right]\left(4⋅\sin^2 \left[\dfrac{x+y}{2}\right]+4⋅\cos^2 \left[\dfrac{x+y}{2}\right]\right)=2a^2$

$4⋅cos^2 \left[\dfrac{x-y}{2}\right]=2a^2$

$cos^2 \left[\dfrac{x-y}{2}\right]=\dfrac{2a^2}{4}$

$⇒cos\left[\dfrac{x-y}{2}\right]=\dfrac{a}{\sqrt{2}}$

Therefore on substituting back to our equation;

$2 \sin \dfrac {x+y}{2} ⋅ \cos \dfrac {x-y}{2}=a$

and

$2 \cos \dfrac{x+y} {2}⋅\cos \dfrac {x-y}{2}=a$

we get;

$2 \sin \dfrac {x+y}{2} ⋅ \dfrac{a}{\sqrt{2}}=a$

$2 \cos \dfrac{x+y} {2}⋅\dfrac{a}{\sqrt{2}}=a$

$⇒\sin \dfrac {x+y}{2}=\dfrac{\sqrt{2}}{2}$

$⇒\cos \dfrac {x+y}{2}=\dfrac{\sqrt{2}}{2}$

$⇒x+y=\dfrac{π}{2}$

Let $x=y$ then $x=y=\dfrac{π}{4}$

$\sin x +\cos x=\sin \dfrac{π}{4} +\cos \dfrac{π}{4}=\sqrt{2}$

but;
$⇒\cos\left[\dfrac{x-y}{2}\right]=\dfrac{a}{\sqrt{2}}$

since $x=y$ then;

$1=\dfrac{a}{\sqrt{2}}$

$⇒a=\sqrt{2}$

Therefore;

$\sin x + \cos x= a$

That is fine, but where am I mistaken in my solution in post #5 above?

You get $cos\left[\dfrac{x-y}{2}\right]=\dfrac{a}{\sqrt{2}}$.

But I get $cos\left[\dfrac{x-y}{2}\right]=0$.

We are both using correct math. But one of us has got to be wrong.

I agree it looks like I am. I just want to know where.

I paste my solution again. Please let me know when you get the time.

 

[brotherbobby]

View attachment 317015
Fine. Let us investigate 　
$$x-y=\pi$$
$$a=\sin x+\sin y=\sin (\pi+y)+\sin y = -\sin y + \sin y =0$$
but the problem says $a \neq 0$. So we should avoid it and go to another possibility that you wrote after OR in your last line. This was a repeat of post #8.

Correct. Please carry on.

 

[kuruman]

That is fine, but where am I mistaken in my solution in post #5 above?

You get $cos\left[\dfrac{x-y}{2}\right]=\dfrac{a}{\sqrt{2}}$.

But I get $cos\left[\dfrac{x-y}{2}\right]=0$.

We are both using correct math. But one of us has got to be wrong.

I agree it looks like I am. I just want to know where.

I paste my solution again. Please let me know when you get the time.

View attachment 317014

You have $$\begin{align} & 2\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)=a \nonumber \\ & 2\cos\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right)=a \nonumber
\end{align}$$If you divide the top equation by the bottom, you get $$\tan\left(\frac{x+y}{2}\right)=1 \implies \frac{x+y}{2}=\frac{\pi}{4}.$$ Explore where that takes you.