The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Translating Trig to Algebra; Proving Trig Identities

Date: 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)


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.


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:

     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:

     1 / |?
      /  |
     /   |
   A   x  C

Now we can use the Pythagorean theorem to fill in the missing side:

     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. 

    y = sin(arccos(x))

then this means the same as

    y = sin(A)  and  A = arccos(x)


    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 
Associated Topics:
High School Trigonometry

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.