Hey guys, I have a problem:
Knowing that
$\displaystyle sin \alpha -cos \alpha = \frac{1}{5}$ and $\displaystyle \frac{ \pi }{4} < \alpha < \frac{ \pi }{2}$, find $\displaystyle tan \alpha$
Thanks in advance!
i am inclined to square both sides:
(sin(a) - cos(a))^2 = 1/25
sin^2(a) + cos^2(a) - 2sin(a)cos(a) = 1/25
1 - 2sin(a)cos(a) = 1/25
24/25 = 2sin(a)cos(a) = sin(2a)
and use the fact that tan((1/2)arcsin(x)) = sin(arcsin(x))/[1 + cos(arcsin(x))]
= x/(1 + √(1-x^2))
although this is the "long way 'round" it has the advantage of not relying on knowing the answer before you solve the problem
Except we are all expected to "know" the most basic fact-iods about triangles (and I would suggest that the appearance of "magic" numbers is indicative of what solution method is expected), and such knowledge is more common than that of some trig-identities.
Also the term absurd comes to mind when people suggest a more complex solution method when an elementary one is staring them in the face.
Ohh.. and by the way you still have a horrible bit of algebra to do...
Contrary to what many people think one of the purposes of mathematics is to make simple that which looks complicated (and not the other way around unless for a good pedagogical reason, which I see not here)
CB
is mathematics a process of applying what we know, to discover what we don't, or a game of "guess the answer!"?
if it is the latter, then i concur with you, and once one knows how to google "wolframaplha", should immediately abandon doing things "the hard way".
if mathematics is merely a test of what trivia you have picked up, we are both wasting our time here.
(EDIT: pfft! i know, the algebra is horrid....can you not see the enormous bulge in my cheek caused by the internal protuberance of my tongue? yes 4/5 - 3/5 = 1/5, a fact that some egyptian engineer no doubt put to good use in ensuring that pyrmaid didn't get built crooked).