What is the Sum of Angles a and b in This Trigonometry Problem?

AI Thread Summary
The discussion focuses on solving a trigonometry problem involving two equations: sin(a) + cos(b) = -1/2 and cos(a) + sin(b) = √3/2, with specified ranges for angles a and b. The approach suggested involves squaring both equations and adding them to eliminate variables, leading to the expression sin(a + b). This method simplifies the problem and allows for the determination of sin(a + b) directly from the given equations. The participant confirms that this approach successfully leads to the solution, expressing gratitude for the guidance. The discussion concludes with a resolution to the problem using the outlined method.
Helly123
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Homework Statement



##\sin a + \cos b## = ##\frac{-1}{2}##
##\cos a + \sin b## = ##\frac{\sqrt 3}{2}##

0 < a < ##\pi/2##
##\pi/2## < b < ##\pi##

a + b = ? By calculating sin (a+b)

Homework Equations

The Attempt at a Solution


I tried :
##\sin a + \cos b =
2sin\frac{(a+b)}{2}cos\frac{(a-b)}{2} = -\frac{1}{2}##

##\cos a + \sin b =2sin\frac{(a+b)}{2}cos\frac{(b-a)}{2} = \frac{\sqrt3}{2}##

I tried to multiple it by ##\sqrt2/2##
##\sin a \cos 45 + \sin 45 \cos b = -\frac{1}{2}\frac{\sqrt 2}{2}##
##\sin 45 \cos a + \sin b \cos 45 = \frac{\sqrt3}{2}\frac{\sqrt 2}{2}##

##\sin a + \cos b = -\frac{1}{2} = \sin 210 = \sin 330 ##

None of this steps get me a clue to find a + b. Can i get a clue?
 
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What is your usual approach when solving two simultaneous equations? Don't you try to eliminate one variable?
 
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Square the two equations and add.

You'll get ##\sin a \cos b + \cos a \sin b## on the left hand side along with other terms. Other terms will reduce to 1. ##\sin a \cos b + \cos a \sin b = \sin (a+b)##. You're done.
 
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Work at it backwards. The problem asks you to find a + b by first finding sin(a + b). What is sin(a + b) ? Then see if you can get that from the given equations
 
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Wrichik Basu said:
Square the two equations and add.

You'll get ##\sin a \cos b + \cos a \sin b## on the left hand side along with other terms. Other terms will reduce to 1. ##\sin a \cos b + \cos a \sin b = \sin (a+b)##. You're done.
It worked. Thanks
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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