Prove: $\sin P+\sin Q> \cos P+\cos Q +\cos R$ | Trig Challenge

[anemone]

Let $P,\,Q,\,R$ be the angles of an acute-angled triangle. Prove that $\sin P+\sin Q> \cos P+\cos Q +\cos R$.

 

[anemone]

Hint:

Spoiler

In any acute-angled triangle, we have $\angle P+\angle Q>\dfrac{\pi}{2}$ and hence $\sin P>\cos Q$.

Further hint:

Spoiler

$\therefore 1-\sin P<1-\cos Q$

 

[anemone]

Solution of other:

Spoiler

In any acute-angled triangle, we have

$\angle P+\angle Q>\dfrac{\pi}{2}$

[TABLE="class: grid, width: 600"]
[TR]
[TD]$\sin P>\sin\left(\dfrac{\pi}{2}-Q\right)=\cos Q$

$1-\sin P<1-\cos Q$[/TD]
[TD]or[/TD]
[TD]$\sin Q>\sin\left(\dfrac{\pi}{2}-P\right)=\cos P$

$1-\sin Q<1-\cos P$[/TD]
[/TR]
[/TABLE]

Multiplying the above two inequalities gives

$(1-\sin P)(1-\sin Q)<(1-\cos P)(1-\cos Q)$

$1-(\sin P+\sin Q)+\sin P\sin Q<1-(\cos +\cos Q)+\cos P\cos Q$

$\begin{align*}\sin P+\sin Q&>\cos P +\cos Q+\sin P\sin Q-\cos P\cos Q\\&=\cos P +\cos Q-\cos(P+Q)\\&=\cos P +\cos Q-\cos(\pi-R)\\&=\cos P +\cos Q+\cos R\end{align*}$