<p>I'm taking precal right now and there are a few homework problems that I can't seem to figure out: </p>
<p>1) solve for x:
cos^-1 (x) = tan^-1 (x)<br>
(inverse cosine of x is equal to inverse tangent of x) </p>
<p>2) Show that </p>
<p>tan^-1 (x) + tan^-1 (y) = tan^-1 (x+y)/(1-xy) </p>
<p>for x > 0<br>
y > 0<br>
xy < 1</p>
<p>(inverse tangent of x plus inverse tangent of y equals inverse tangent of (x+y) divided by (1-xy))</p>
<p>Thanks!</p>
<p>1)</p>
<p>Take tan of both sides</p>
<p>Sin(Acos)/Cos(Acos)=Tan(Atan)
SinAcos/x=x</p>
<p>Use the Pythagorean identity:</p>
<p>Sqrt(1-Cos^2Acos)=x^2
Sqrt(1-x^2)=x^2</p>
<p>Solve by squaring and using quad form:</p>
<p>x^4+x^2-1=0
x^2=(-1+-Sqrt5)/2</p>
<p>neg solution is extraneous (we're dealing with reals here):</p>
<p>x=Sqrt(-1+Sqrt5)/2</p>
<p>2) I can't seem to solve this one algebraically. It reduces to a/b=Tan(Arctan(a)/b) where a=x+y and b=1-xy. This is equivalent to Atan(a/b)=Atan(a)/b.</p>
<p>Possibly there's a super simple solution, but I cannot see it.</p>
<p>For these times of horny questions, and the solve questions in general, I recommend that if you cannot solve immediately, then plug numbers using the calculator.</p>
<p>SAT II is not a test of mathematical ability. SAT II is a test of 1) understanding the problem and 2) plugging into calculator. Evidently, one can also use skill to solve the probs rather than calculator, but unless you are truly proficient at elementary precalc mathematics, using the calculator is the sure-fire way to go. It's also a lot easier to use the calculator.</p>
<p>In no way am I scorning your mathematical skill here, but this is the truth. I've won a good share of math contests in HS, and for me, it was still 10x easier to use the calculator. Even if one is truly a wizard, the algebraic method leaves room for error. The calculator does not, except in pressing the right keys (and I think we all agree than keyboard input is a hell a lot easier than solving inverse trig with two vars).</p>
<p>sphoenixee</p>
<p>^^
2) sphoenixee, taking tan of both sides works here too; you just apply this identity:
tan(a + b) = (tan a + tan b) / (1 - (tan a)(tan b))
(it's easily derived from left to right substituting sin/cos for tan).</p>
<p>If arctan x = a and arctan y = b,
then tan a = x and tan b = y.
We can rewrite now<br>
tan^-1 (x) + tan^-1 (y) = tan^-1 (x+y)/(1-xy) as
a + b = arctan ((x+y)/(1-xy)).</p>
<p>tan(a + b) = tan (arctan ((x+y)/(1-xy)))
(tan a + tan b) / (1 - (tan a)(tan b)) = (x+y)/(1-xy)<br>
(x+y)/(1-xy) = (x+y)/(1-xy) ------> done.</p>
<p>=============
1).
This question allows an interesting geometric interpretation on the unit circle (I am not saying its a better way to solve).</p>
<p>Let arccos x = arctan x = a, then
cos a = x, tan a = x</p>
<p>If point A on the circle in the first quadrant corresponds to this angle "a", and point A projects to point C on the x-axis, then in the right triangle OCA
OA = 1 and OC = x,
so CA = sqrt(1 - x^2).
tan a = tan (<aoc) =="" ac="" oc="" tan="" a="sqrt(1" -="" x^2)="" x.="" going="" back="" to="" x="sqrt(1" x,="" x^2="sqrt(1" --=""> the same equation you got.</aoc)></p>
<h1>The end of the interpretation.</h1>
<p>Just to finish:
Substitute t = x^2
t^2 = 1 - t
t^2 + t - 1 = 0
t = (-1 + sqrt 5) / 2, since x^2>0
x = +-sqrt ((-1 + sqrt 5) / 2).
arccos x is in the first and second quadrants,
arctan x is in the first and forth quadrants,
so
arccos x = arctan x = a has to be in the first quadrant, where x > 0:
x = sqrt ((-1 + sqrt 5) / 2).</p>