Saurav B.

asked • 08/28/14

if cos(A+B)+sin(A-B)=0and tanB=1/2006.find tanA

if cos(A+B)+sin(A-B)=0and tanB=1/2006.find tanA

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Phillip R. answered • 08/28/14

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cosAcosB - sinAsinB + sinAcosB - sinBcosA = 0
 
cosA(cosB - sinB) + sinA(cosB - sinB) = 0
 
(cosB - sinB) (cosA + sinA) = 0
 
Therefore cosB - sinB = 0     or     cosA + sinA = 0
 
cosB = sinB               cosA = -sinA
 
tanB = 1                   tanA = -1
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Stan Z.

Regarding the answer provided by Phillip R.  You are showing step by step a method you chose to take through to completion.  Suggesting a method and providing a hint or two what could be done may be more instructive; however, you end by stating TanB=1 when the problem states as a given TanB=1/2006.  This must be confusing to Saurav.  Another path could have shown that it mattered not what TanB's value is and if the problem insists it be 1/2006 than albeit as the answer regardless for TanA is as you've shown to be -1.
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08/28/14

Phillip R.

I have no idea what your point is. Furthermore it is impossible for tanB to equal that fraction under the condition stated. Furthermore my way of helping the students is not your concern. 
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08/28/14

Stan Z.

Hi Phillip:
 
You are absolutely right and I totally apologize if I offended you in any way. Your way is not my concern and I hope you accept my apology. 
 
In the spirit of the problem may I point out your therefore statement that says CosB-sinB=0 OR cosA + sinA =0.  We all know that extraneous roots can crop up.  The only valid solution under the criteria given is cosA + SinA =0 leading to what was asked, TanA=-1.  And if TanA=-1 then TanB can take on ANY value even 1/2006.
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08/28/14

Phillip R.

I must correct you. I did the busy work but left the conclusion to the student. The operative word is OR. TanB can = 1 and when it does TanA can take on any value. Similarly when TanA = -1, TanB can have any value including 1/2006. You are not correct to say TanA = -1 regardless of the value of TanB because when TanB =1, TanA is not necessarily -1. And so the essence of OR. TanA can have any value when TanB = 1. Because it equals a number other than 1, TanA must take the value = -1. This reasoning was left for the student. The busy work that showed was not the important part of the learning experience. 
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08/29/14

Stan Z. answered • 08/28/14

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Hi Saurav:
 
This problem involves you knowing the sum angle formulas.  That is Cos(A+B) = CosACosB - SinASinB and
 
sin(A+B) = SinACosB + CosASinB
 
You also need to know that TanA = SinA/CosA.  With this information and the fact that states Cos(A+B) + Sin(A-B) =0 you should be able to show that TanA =-1.
 
Once you establish that the Cos(A+B) + Sin(A-B) = 0 then dividing first by the Cos(A) then dividing by Cos(B) by making the assumption that neither the cos(A) nor the Cos(B) is zero, and doing some algebra steps, it should become apparent that TanA = -1 regardless of what the tanB is!
 
 
 
 
 
 
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