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

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Homework Help Overview

The problem involves two simultaneous trigonometric equations: sin a + cos b = -1/2 and cos a + sin b = √3/2, with specified ranges for angles a and b. The goal is to determine the sum of angles a and b.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods for solving the equations, including elimination of variables and squaring the equations. There is also a suggestion to work backwards from the desired outcome of finding sin(a + b).

Discussion Status

Some participants have offered guidance on potential approaches, such as squaring the equations and adding them. Others have questioned the original poster's methods and suggested alternative strategies. The discussion reflects a mix of interpretations and attempts without reaching a consensus.

Contextual Notes

The original poster expresses difficulty in progressing towards a solution and seeks a clue, indicating a potential gap in understanding or application of trigonometric identities.

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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