Question

I'm just curious on how to do this:D Thanx!!:)

Answer 1

I can think of a few ways you could try...
One way would be to convert to sines and cosines and then use their translation identities: \( \sin \left( \theta - \dfrac\pi 2 \right) = - \cos \theta\) and \( \cos \left( \theta - \dfrac\pi 2 \right) = \sin \theta \).

Another way might be using the angle addition / subtraction identities for either cotangent / tangent or sine / cosine (if using those above seems redundant).

Answer 2

Ok. I'm not sure how I would work this out though.

Answer 3

Have you tried anything so far that may or may not have worked out so far?

Answer 4

No. But I really wanna know. I'm done with school, but it was online. And I pretty much didn't learn anything in the course.

Answer 5

For verifying trig identities, I would always have some base rules that I know are true:
sin, cos, csc, sec, tan, cot; and how they relate
sin^2 x + cos^2 x = 1 (this is the basis for many many others)
addition and subtraction formulas for sine and cosine
sin(a+b) = sin a cos b + sin b cos a
cos(a+b) = cos a cos b - sin a sin b
and then just start playing with it, having the goal of matching the left to the right or getting a known identity again in the end.

Answer 6

Ok!! I think I got it!!:) Thanx for the tipsxD!!

Answer 7

No problem!

As an example, if you wanted to know the angle sum for tangent, you just have to figure out:
\( \begin{align} \tan (a+b) &= \dfrac{\sin (a+b)}{\cos (a+b)} \\ &= \dfrac{\sin a \cos b + \sin b \cos a}{\cos a \cos b - \sin a \sin b} \end{align} \)
The tangent angle sum identity is usually written in terms of tangent. We could divide both top and bottom of this fraction by cos a cos b, then, since that either cancels a numerator cosine or creates tangent from sine.
\( \tan (a+b) = \dfrac{\tan a + \tan b}{1 + \tan a \tan b} \)

And of course we could do the same for cotangent with roughly the same strategy, an then we just calculate for this particular example cot(theta - pi/2) = -tan theta