Translating Trig to Algebra; Proving Trig IdentitiesDate: 08/09/2002 at 20:40:59 From: Cornelius Willoway Subject: Trig, trig inverse, trig identities Write each expression as an algebraic expression in x, free of trig or inverse trig functions. sin(arccos x) tan(arcsin x) cos(arccot x) sec(arctan(x+2)) I know you have to make a triangle and assign values but how to get those values I am clueless. Prove these identities sec^2 x + csc^2 x = sec^2 x (csc^2 x) sin^4 x + 2sin^2 x (cos^2 x) +cos^4 x = 1 I don't know where to begin with these. I don't know any identities that turn addition into multiplication. Also, the 4th power confuses me in the second one. Please solve these so I can see the steps you used so that I may try to apply them to some other problems I have. Cornelius Date: 08/09/2002 at 22:49:54 From: Doctor Peterson Subject: Re: Trig, trig inverse, trig identities Hi, Cornelius. Let's look at the first expression, sin(arccos x). Suppose we have an angle whose cosine is x (that is, an arccos of x). If we assume to start with that it is acute, we can draw a triangle like this, where the angle whose cosine we know will be A: B + /| c / |a / | / | +----+ A b C The cosine of A is b/c. If we want that to be equal to x, we might as well make the denominator 1 and the numerator x, so that b/c = x/1: B + /| 1 / |? / | / | +----+ A x C Now we can use the Pythagorean theorem to fill in the missing side: B + /| 1 / |sqrt(1-x^2) / | / | +----+ A x C Now we want to find the sine of A. What is that? It's a/c, which in this figure becomes sqrt(1-x^2). To finish the work properly, unless you have been told that all angles are acute, we should check out the more general case. First, what is the range of the arccos function? 0 < arccos(x) < pi. So if x is negative, the arccos will be an angle in the second quadrant. Since the sine of such an angle will still be positive, we don't need to change our answer in this case. Having done this using a picture, it would be a good idea to do it a little more carefully in algebraic form - pictures can be deceiving. If y = sin(arccos(x)) then this means the same as y = sin(A) and A = arccos(x) or y = sin(A) and cos(A) = x and 0 < A < pi Knowing that sin^2(A) + cos^2(A) = 1 this gives y^2 + x^2 = 1 so that y = +/- sqrt(1-x^2) The range of A tells us that the sine is positive, and we can always choose the positive sign in the answer. This agrees with what we found using the picture. And the picture is all you need if you are restricted to acute angles and positive trig functions. >Prove these identities > >sec^2 x + csc^2 x = sec^2 x (csc^2 x) > >sin^4 x + 2sin^2 x (cos^2 x) +cos^4 x = 1 I almost always convert everything to sines and cosines first, because I know more identities using those than secants and cosecants. Then if there are fractions, I multiply everything by the common denominator to make it neater. Once you do that, you will be done with the first. For the second, the Pythagorean identity relates squares of sines and cosines, so I would try writing all the squares of cosines in terms of the sines using that. The 4th power is just (cos^2(x))^2, which becomes (1 - sin^2(x))^2. Expand all the squares of binomials, and you will have something you can work with. If you need more help, please write back and show me how far you got on these or other problems. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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