SOLUTION: If sin(x)+cos(x) = -1/5, and 3pi/4 ≤ x ≤ pi, find the value of cos(2x)

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Question 1153160: If sin(x)+cos(x) = -1/5, and 3pi/4 ≤ x ≤ pi, find the value of cos(2x)
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
If sin(x)+cos(x) = -1/5, and 3pi/4 ≤ x ≤ pi, find the value of cos(2x)


Clear of fractions by multiplying thru by 5



Use identity 
              



Isolate the term with the square root



Square both sides:







divide through by 2



Factor the right side



Use the zero-factor property

  

   

   

Since we are given 



which means x is in QII where the sine is positive,
we can eliminate the negative value for the sine.



Use identity
              













Edwin

Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
It can be done in much simpler manner.


Start from the given equation

    sin(x) + cos(x) = .


Square both sides

    sin^2(x) + 2*sin(x)*cos(x) + cos^2(x) = .


Replace  sin^x) + cos^2(x) by 1

    1 + 2*sin(x)*cos(x) = 


Replace 2*sin(x)*cos(x) by sin(2x)

    sin(2x) =  - 1,   or   sin(2x) = .


Hence, cos^2(2x) =  =  =  =  = .


Since the angle (2x) is in QIV, cos(2x) is positive.


Hence,  cos(2x) = positive square root of  =  = .


The proof is completed.


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