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Posts:
153
Registered:
10/3/07


Re: Proving trigonometric identities
Posted:
Jan 21, 2009 10:48 PM


On Jan 21, 7:32 pm, Albert <albert.xtheunkno...@gmail.com> wrote: > Moving away just a little away from proof of trigonometric identities > and onto solving 'simple' trigonometric equations: > > When I get questions like: > Find solutions in the range 0 degrees <= x <= 360 degrees for these > trigonemetric equations: > (a) sin x = 1/2; > (b) ... > ... > > Do I figure out one answer (say 30 degrees) and then think: there > might be another answer and if so, the sin of it better be positive as > well, which would mean it'd be in the 2nd quadrant and figure out 150 > degrees? Is that the way to do these problems because I don't have any > worked examples whatsoever. > > I don't have any worked examples for questions like these either: > (a) Write sin(theta) and tan(theta) in terms of cos(theta) when theta > is in the first quadrant > (b) If cosA = 9 / 41, and A is in the first quadrant, find tanA and > cosecA. > > I know that you can get a calculator and solve for A using inverse > cosine immediately, but what is the real intention of this question? > The trig identity tan(theta) = sin(theta) / cos(theta) can be re > arranged for finding a value for sin(theta) but i) what is the > significance of which quadrant theta is in (and) how do I find tanA in > part b?
Hello,
Let us find sin(theta).
Recall that (sin x)^2 + (cos x)^2 = 1 for all x.
If cos(theta) = 9/41, then (sin(theta))^2 = 1 (9/41)^2.
It so happens that 9^2 + 40^2 = 41^2, so 1  (9/41)^2 = (40/41)^2. But even if things did not turn out so nicely, we could continue.
Thus sin(theta) = 40/41. This is where we use the fact that theta is in the first quadrant. If instead theta were in the fourth quadrant, we would have sin(theta) = 40/41.
Anyway, now that we have sin(theta), anything else can be calculated. For example, tan(theta( = 40/9.

