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the question begin with prove that

$$(\sin A+\cos A)^2 +(\sin B+\cos B)^2+(\sin C+\cos C)^2=\sin 2A-\sin 2B-\sin 2C-1$$


Given $A+B+C = 180^\circ$ it is proved that $$(\sin A+\cos A)^2 +(\sin B+\cos B)^2+(\sin C+\cos C)^2=-4\sin A \cos B \cos C-1$$

Last, it asked us to find the range of $\cos B\cos C$ if $A=90^\circ$

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1 Answer 1

Note that $\widehat{B}=90^{\circ}-\widehat{C}$.

So, $\cos\widehat{B}\cos\widehat{C}=\cos\widehat{B}\cos(90^{\circ}-\widehat{B})=\cos\widehat{B}\sin\widehat{B}=\frac{1}{2}\sin(2\widehat{B})$.

That means that $\cos\widehat{B}\cos\widehat{C}\in\left[-\frac{1}{2};\frac{1}{2}\right]$.

(Supposing that $\widehat{B}$ can have any value)

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