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Topic: Proving trigonometric identities
Replies: 32   Last Post: Jan 27, 2009 1:59 AM

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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
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.

Date Subject Author
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 Angus Rodgers
1/20/09 Guest
1/20/09 Angus Rodgers
1/20/09 Albert
1/20/09 Angus Rodgers
1/20/09 matt271829-news@yahoo.co.uk
1/21/09 Albert
1/21/09 matt271829-news@yahoo.co.uk
1/23/09 Albert
1/23/09 Angus Rodgers
1/23/09 Angus Rodgers
1/23/09 Passerby
1/23/09 Dave Dodson
1/24/09 Albert
1/24/09 Angus Rodgers
1/24/09 Albert
1/26/09 Albert
1/26/09 Driveby
1/26/09 Albert
1/26/09 A N Niel
1/27/09 Albert