Doubt from theory of equations


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Now cosA+cosB+cosC less than 3/2
If A=B=C then cosA+cosB+cosC=3/2
So cosA+cosB+cosC+cosC(k-1)
3/2+1/2(k-1)= 1+1/2(k)
So minimum value greater than1+1/2×1/2 =5/4
Wait i am trying

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Maximum value of cosA+cosB+cosC=3/2
Now 3/2-cosC+kcosC
=1+k/2 since k is greater than 1/2
Then (1+k/2) is greater than (1+1/2×1/2)=5/4
So in my opinion there is no maximum value but
But the given expression is greater than 5/4

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Given A+B+C=\pi and Let u=\cos (A)+\cos B+k\cos(C).

\displaystyle \Longrightarrow u=2\cos\bigg(\frac{A+B}{2}\bigg)\cos\bigg(\frac{A-B}{2}\bigg)+k\cos C

\displaystyle \Longrightarrow u=2\sin \frac{C}{2}\cos\bigg(\frac{A-B}{2}\bigg)+k\bigg(1-2\sin^2 \frac{C}{2}\bigg).

\displaystyle \Longrightarrow 2k\sin^2 \frac{C}{2}-2\sin \frac{C}{2}\cos\bigg(\frac{A-B}{2}\bigg)+u-k=0

For real Roots, Its Discriminant \mathcal{D}\geq 0.

So we have \displaystyle 4\cos^2\bigg(\frac{A-B}{2}\bigg)-8k(u-k)\geq 0

So finally we have \displaystyle 8k(u-k)\leq 4\cos^2\bigg(\frac{A-B}{2}\bigg)\leq 4.

\displaystyle \Longrightarrow 2k(u-k)\leq 1\Longrightarrow \color{brown}u\leq \frac{1}{2k}+k.

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Let A=u+v, B=u-v, then C=\pi - 2u

Then given expression = \cos (u+v)+\cos (u-v)-k\cos 2u =2\cos u \cos v -2k \cos^2 u +k

\le 2 \cos u - 2k \cos^2 u +k = \dfrac{1}{2k}+k - \left(\sqrt {2k} \cos u - \dfrac{1}{\sqrt{2k}} \right)^2 \le k+\dfrac{1}{2k} with equality when v=0 and \cos u = \dfrac{1}{2k} which is a valid equation since \dfrac{1}{2k}<1