Solving Trig Problem: sin^2000(x) + cos^2000(x)=1 - Kirstin

[kirstin.17]

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.

 

[arildno]

You KNOW that [itex]\sin^{2}(x)+\cos^{2}(x)=1[/tex]<br /> Set $a=\sin^{1998}(x), b=\cos^{1998}(x)$<br /> and your equation can be written as:<br /> $$a\sin^{2}(x)+b\cos^{2}(x)=1$$<br /> <br /> Subtract the first from the second, yielding:<br /> $$(a-1)\sin^{2}(x)+(b-1)\cos^{2}(x)=0$$<br /> What can you conclude about this expression?[/itex]

 

[kirstin.17]

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 $$(a-1)\sin^{2}(x)=0$$ and $$(b-1)\cos^{2}(x) =0$$, or where $$(a-1)\sin^{2}(x)=-(b-1)\cos^{2}(x)$$.

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]

Indeed you are correct.

 

[mydarshankumar]

......
good ans