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

AI Thread Summary
The discussion focuses on finding the value of k² in a trigonometric context, where initial calculations suggest k² = 2, but the textbook provides a different solution of k² = 4ab/(a² + b²). Participants clarify the steps leading to the correct factorization and simplification of the equations, emphasizing the importance of including terms like cos²(C/2) in the calculations. A key point is the realization that proper expansion and division by (a+b)² are necessary for accurate results. The conversation highlights the need for careful step-by-step explanations to avoid confusion in mathematical problem-solving.
chwala
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Homework Statement
See attached.
Relevant Equations
Trigonometry
1697019239760.png


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?
 
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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
 
anuttarasammyak said:
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.
 
Last edited:
chwala said:
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 said:
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!
 
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)##
 
Last edited:
SammyS said:
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...
 
chwala said:
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:

chwala said:
##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)##
 
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