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What is the maximum value of $2(\sin A + \sin B) - 2\cos C$ in a triangle $ABC$? Can anyone help?

ronal
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1 Answers1

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Let $y=2(\sin A+\sin B)-2\cos C=4\cos\dfrac C2\cos\dfrac{A-B}2-2\left(2\cos^2\dfrac C2-1\right)$

$$\iff4\cos^2\dfrac C2-4\cos\dfrac C2\cos\dfrac{A-B}2+y-2=0$$

Now as $\cos\dfrac C2$ is real, the discriminant must be $\ge0$

i.e., $$\left(4\cos\dfrac{A-B}2\right)^2\ge16(y-2)$$

$$\iff y-2\le\cos^2\dfrac{A-B}2\le1$$ which occurs if $A=B$ as $0<A,B<\pi$

consequently $\cos\dfrac C2=\dfrac12\iff C=\dfrac{2\pi}3$ as $0<C<\pi$

  • Thank you so much! – ronal Apr 22 '18 at 03:31
  • @ronal, See also : https://math.stackexchange.com/questions/952893/cos-a-cos-b-cos-c-leq-frac18 and https://math.stackexchange.com/questions/949530/in-triangle-abc-show-that-1-lt-cos-a-cos-b-cos-c-le-frac-32 and https://math.stackexchange.com/questions/2648521/trigonometric-inequality-sina-sinb-cosc-le-frac32 – lab bhattacharjee Apr 22 '18 at 03:36