Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $A=90^\circ$

share|cite|improve this question

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)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.