Trigonometric inequalities in a triangle

Question

What is the proof of $~~\cos^2 A+\cos^2 B+\cos^2 C \leq 1 ~~$ in an acute triangle ?

This will be of help in finding the answer (if such exists) to finding the minimal T in any ∆ ABC when $$T \geq \sin^k A+ \sin^k B+ \sin^k C~ ,~~~~ k \geq 3$$

Answer 1

We have \begin{align} \cos^2A+\cos^2B+\cos^2C&=\frac12(1+\cos2A)+\frac12(1+\cos2B)+\cos^2C\\ &=\frac12(2+\cos2A+\cos2B)+\cos^2C\\ &=\frac12(2+2\cos(A-B)\cos(A+B))+\cos^2C\\ &=1+\cos(A-B)\cos(\pi-C)+\cos^2C\\ &=1-\cos(A-B)\cos C+\cos^2C\\ &=1-\cos C(\cos(A-B)-\cos(\pi-(A+B)))\\ &=1-\cos C(\cos(A-B)+\cos(A+B))\\ \cos^2A+\cos^2B+\cos^2C&=1-2\cos A\cos B\cos C\\ \implies\cos^2A+\cos^2B+\cos^2C&\leq1\\ \end{align}

Answer 2

Use Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$

$$\cos^2A+\cos^2B+\cos^2C-1$$

$$=\cos^2A+\cos(B-C)\cos(B+C)$$

$$=\cos A(-\cos(B+C))+\cos(B-C)(-\cos A)$$

$$=-\cos A(2\cos B\cos C)$$

Answer 3

Let's first assume that $\cos^2A +\cos^2B+ \cos^2C=1$

Let's proceed to solve this

$\cos^2A +\cos^2B+ \cos^2C=1$

$\Rightarrow \cos^2A + \cos^2B - (1−\cos^2C)=0$

$\Rightarrow \cos^2A + cos^2B − sin^2C=0$

$\Rightarrow \cos^2A+(cos(B+C)cos(B−C))=0$

$\Rightarrow \cos^2A+(cos(π−A)cos(B−C))=0$ as $A+B+C = π$

$\Rightarrow \cos^2A−cosAcos(B−C)=0$

$\Rightarrow \cos A(cosA−cos(B−C))=0$

$\Rightarrow cosA(cos(π−(B+C))−cos(B−C))=0 $

$\Rightarrow (−cosA(B+C)−cos(B−C))=0 $

$\Rightarrow cosA(cosA(B+C)+cos(B−C))=0 $

$\Rightarrow −cosA(2cosBcosC)=0 $

$\Rightarrow 2cosAcosBcosC=0 $

$\Rightarrow cosA=0 or cosB=0 or cosC=0 $

$\Rightarrow A=\frac{π}{2} or B=\frac{π}{2} or C=\frac{π}{2} $

With this we can conclude that the triangle ABC will be a right-angled triangle if $\cos^2A +\cos^2B+ \cos^2C=1$.

Now when we say that $\cos^2A +\cos^2B+ \cos^2C \leq 1$ this means that all angles are lesser than $\frac{π}{2}$ which means that the given triangle is Acute angled.

I hope this helped you.