Advanced Math/trig.

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Expert: - 1/13/2009


Question
Hello,

I wondered if you could simplify or give me ideas about how to start simplifying the following:

A = arctan (tan(b) sin [arctan (tan(a) cos(b) ] ).

It's a carpentry formula I have worked out but feels clumsy to use in it's current state.

Thank you very much

Answer
I will assume that a and b are the other two angles in a right triangle.  They will be referred to as A and B.

Let's rewrite the angles as capital letters and change the A in the equation on the left to a D (since A, B, and C are used for the triangle):

D = arctan (tan(B)sin[arctan(tan(A)cos(B)]).  Let's simplify it and drop the parenthesis on the trig functions, so it's
D = arctan (tanB sin[arctan(tanA cosB]).  Also, we'll assume that 'a/b sin...' means '(a over b)*sin ...'.  This makes is simpler to read.

The tanB = ctnA, cosB = sinA, so we have
C = arctan (ctnA sin[arctan(tanA sinA)]), so now everything is in terms of angle A.

So to do all this in Excel and make sure it works, I will reduce the problem from the start.  I used a,b,c as 3,4,5 and got the following:
0.6   sinA   a/c
0.6   cosB   a/c
0.8   cosA   b/c
0.8   sinB   b/c
0.75   tanA   a/b
1.33   tanB   b/a
1.33   ctnA   b/a
0.75   ctnB   a/b

From here, I then manipulated the formula on the right and got the number on the left:
0.500654712   arctan (tan(B) sin[arctan (tan(A) cos(B)])
0.500654712   arctan (ctn(A) sin[arctan (tan(A) sin(A)])
0.500654712   arctan (b/a sin[arctan(a/b a/c)])
0.500654712   arctan (b/a sin[arctan(a²/bc)])
0.500654712   arctan ((b/a)(a²/√(a^4+b²c²)))
0.500654712   arctan [(ba)/√(a^4+b²c²)]

What I finally get is arctan[ba/√(a^4+b²c²)], which looks a whole lot simpler to me.  Look good to you?

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