Trig identity question

  • Thread starter kirstin.17
  • Start date

Main Question or Discussion Point

I'm trying to solve this trig problem:

sin^2000(x) + cos^2000(x) = 1

I'm not sure how to go about it... I tried starting with sin^2(x) + cos^2(x) = 1 and build up to 2000 but I didn't get very far.

Obviously any multiple of pi will be an answer since either sin^2000 or cos^2000 will be 1 and the other will be 0. Are there others as well?

thx
-Kirstin.
 

Answers and Replies

arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,946
130
You KNOW that [itex]\sin^{2}(x)+\cos^{2}(x)=1[/tex]
Set [itex]a=\sin^{1998}(x), b=\cos^{1998}(x)[/itex]
and your equation can be written as:
[tex]a\sin^{2}(x)+b\cos^{2}(x)=1[/tex]

Subtract the first from the second, yielding:
[tex](a-1)\sin^{2}(x)+(b-1)\cos^{2}(x)=0[/tex]
What can you conclude about this expression?
 
Aha..... I would say there are no other solutions other than what I already mentioned.

First note that the last expression you stated is equivalent to the original one we want to solve.

Solutions will occur either where [tex](a-1)\sin^{2}(x)=0[/tex] and [tex](b-1)\cos^{2}(x) =0[/tex], or where [tex](a-1)\sin^{2}(x)=-(b-1)\cos^{2}(x)[/tex].

The first situation has the solutions mentioned earlier.

The second situation does not have any solutions since LHS is always negative (since 0 < a < 1) and RHS is always positive.
 
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,946
130
Indeed you are correct. :approve:
 
......................
good ans
 

Related Threads for: Trig identity question

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
13
Views
11K
  • Last Post
Replies
7
Views
996
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
1
Views
2K
Top