Help with an identity

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hcinic

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Jul 9, 2012
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sin7a-sin5a/cos7a+cos5a=tan a (a=alpha)

I start by combining the numerator to be sin2a which I then put in 2sinacosa for, I thought this could be mistake but was unable to find in book, these terms can be combined, right? Same with denominator, should it be cos12a? If so I am running into a problem on where to go next with cos12a. I also considered addition formulas, but know that is incorrect because sin7a-sin5a is not the same as sin(7a-5a). I know tan a=sin a/cos a, and also tan a = sin2a/1+cos2a which led me to believe I was on the right track by making the numerator sin2a.
 
hcinic said:
sin7a-sin5a/cos7a+cos5a=tan a (a=alpha)

I start by combining the numerator to be sin2a which I then put in 2sinacosa for, I thought this could be mistake but was unable to find in book, these terms can be combined, right? Same with denominator, should it be cos12a? If so I am running into a problem on where to go next with cos12a. I also considered addition formulas, but know that is incorrect because sin7a-sin5a is not the same as sin(7a-5a). I know tan a=sin a/cos a, and also tan a = sin2a/1+cos2a which led me to believe I was on the right track by making the numerator sin2a.
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NO! You are adding/subtracting angles which you can not do. Sin(7a) does equal Sin(2a+5a). What do you know about the sum and difference identies of two angles?
 
By sum and difference identities do you mean addition formulas sin(s+t), sin(s-t), cos(s+t), cos(s-t)?
 
I am familiar with these and am starting to understand that these will be the only way to begin this identity, is this correct? There is no need to mess with tan a (the right side of eq)?
 
hcinic said:
[Considering sum and difference identities] will be the only way to begin [to demonstrate] this identity, is this correct?
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Not necessarily, but that is a good starting point. We need to be careful when using words like "only" in math; there are often different approaches to most problems.


There is no need to mess with [the right side]?
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You are free to do whatever you like, on your scratch paper. The acceptableness of what you turn in depends upon your instructor. Some instructors allow students to manipulate both sides, when proving an identity; others do not allow this. (I think that it's a moot point because the latter may always be rewritten as the former.)


By the way, this is what you typed, but I've added some potential grouping symbols:

hcinic said:
[sin7a - sin5a]/[cos7a + cos5a] = tan a
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Without those grouping symbols, your typing means this:

sin(7a)−sin(5a)cos(7a)+cos(5a)=tan(a)\displaystyle sin(7a) - \frac{sin(5a)}{cos(7a)} + cos(5a) = tan(a)
 
Thanks, new to forum and still learning. I know it is very easy to get confused by entering in the wrong information. I still am stuck though with starting with the addition formulas. Is there another way to demonstrate?
 
Nice...thank you subhotosh khan. Haven't tried that way yet, but it already makes perfect sense! :D
 
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