Find the value of ##k^2## in the problem involving trigonometry

[chwala]

Homework Statement

See attached.

Relevant Equations

Trigonometry

In my working i have,

...

$\cos C = 2\cos^2 \dfrac{1}{2} C -1$

$c^2= a^2+b^2-2ab(2\cos^2 \dfrac{1}{2} C-1)$

$c^2= a^2+b^2+2ab(1-2\cos^2 \dfrac{1}{2} C)$

$c^2= (a+b)^2 (1-2\cos^2 \dfrac{1}{2} C)$
Now from here, $k^2 =2$ but text gives different solution. I am still checking this...am i missing something guys?

 

[anuttarasammyak]

From the third line
$$c^2=(a+b)^2 (1 - 4\frac{ab}{(a+b)^2}\cos^2 \frac{C}{2})$$
So
$$k=\frac{\sqrt{ab}}{\frac{a+b}{2}} \leq 1$$

 

[chwala]

From the third line
$$c^2=(a+b)^2 (1 - 4\frac{ab}{(a+b)^2}\cos^2 \frac{C}{2})$$
So
$$k=\frac{\sqrt{ab}}{\frac{a+b}{2}} \leq 1$$

Ok nice one, i can see that

$(a+b)^2\left[\dfrac{(a+b)^2-4ab}{(a+b)^2}\right]≡a^2+b^2+2ab(1-2\cos^2 \dfrac{1}{2} C)$

but the text solution is $k^2= \dfrac{4ab}{a^2+b^2}$ and not $k^2= \dfrac{4ab}{(a+b)^2}$ as you've shown.

I hope i checked my textbook correctly...will have access to it later.

 

[SammyS]

Ok nice one, i can see that

$(a+b)^2\left[\dfrac{(a+b)^2-4ab}{(a+b)^2}\right]≡a^2+b^2+2ab(1-2\cos^2 \dfrac{1}{2} C)$

but the text solution is $k^2= \dfrac{4ab}{a^2+b^2}$ and not $k^2= \dfrac{4ab}{(a+b)^2}$ as you've shown.

I hope i checked my textbook correctly...will have access to it later.

You have left out the cos² on the left hand side of that first equation. It should read:

$\displaystyle \quad\quad (a+b)^2\left[\dfrac{(a+b)^2-4ab\,\cos^2(C/2)}{(a+b)^2}\right] = \dots$

This is consistent with @anuttarasammyak's result and simplifies to:

$\displaystyle \quad\quad (a+b)^2\left[1-\dfrac{(2ab)\,2\cos^2(C/2)}{(a+b)^2}\right]$ ,

which can easily be compared to the 2nd or 3rd line of your OP.

 

[chwala]

Ok nice one, i can see that

$(a+b)^2\left[\dfrac{(a+b)^2-4ab}{(a+b)^2}\right]≡a^2+b^2+2ab(1-2\cos^2 \dfrac{1}{2} C)$

but the text solution is $k^2= \dfrac{4ab}{a^2+b^2}$ and not $k^2= \dfrac{4ab}{(a+b)^2}$ as you've shown.

I hope i checked my textbook correctly...will have access to it later.

@anuttarasammyak you're correct. Cheers!

 

[chwala]

Was good question mixed me up a bit. Wah! Expand with $2ab$ first, then factorize to have $(a+b)^2$ then divide each term by $(a+b)^2$ and multiply whole by $(a+b)^2$.
Will post later once I get hold of laptop.

What i was missing was:
...

$a^2+b^2+2ab(1-2)$

on expanding we get;

$a^2+b^2+2ab-4ab=((a+b)^2 -4ab)$

then divide each term by $(a+b)^2$ and multiplying by $(a+b)^2$ realizes,

$= \left(1-\dfrac{4ab}{(a+b)^2}\right)×(a+b)^2=(a+b)^2\left(1-\dfrac{4ab}{(a+b)^2}\right)$

 

[chwala]

You have left out the cos² on the left hand side of that first equation. It should read:

$\displaystyle \quad\quad (a+b)^2\left[\dfrac{(a+b)^2-4ab\,\cos^2(C/2)}{(a+b)^2}\right] = \dots$

This is consistent with @anuttarasammyak's result and simplifies to:

$\displaystyle \quad\quad (a+b)^2\left[1-\dfrac{(2ab)\,2\cos^2(C/2)}{(a+b)^2}\right]$ ,

which can easily be compared to the 2nd or 3rd line of your OP.

good but that is not really where my problem is though ...my problem is on the factorisation bit...

 

[SammyS]

good but that is not really where my problem is though ...my problem is on the factorisation bit...

Maybe that was your problem, but in the threads you post, it's often difficult to tell where you're having difficulty, because all too often you skip steps and/or do not explain what you're doing.

For instance, in the OP of this thread you have:

$c^2= a^2+b^2+2ab(1-2\cos^2 \dfrac{1}{2} C)$

$c^2= (a+b)^2 (1-2\cos^2 \dfrac{1}{2} C)$

Smaller steps give:

$\displaystyle \quad\quad c^2= a^2+b^2+2ab(1-2\cos^2 (C/2)\, )$

$\displaystyle \quad\quad c^2= a^2+b^2+2ab-4ab\cos^2 (C/2)$

$\displaystyle \quad\quad c^2= (a+b)^2-4ab\cos^2 (C/2)$

$\displaystyle \quad\quad c^2= (a+b)^2\left(1-\dfrac{4ab\cos^2 (C/2)}{(a+b)^2}\right)$