# Math Help - range of trigonometric function

1. ## range of trigonometric function

i have confusion in finding range of sin^4x + cos^2x.
I have replace cos^2x as sin^2x -1 and let sin^2x as t.
It becomes then t^2-t+1. Now how can i proceed?

2. ## Re: range of trigonometric function

Originally Posted by ohm
i have confusion in finding range of sin^4x + cos^2x.
I have replace cos^2x as sin^2x -1 and let sin^2x as t.
It becomes then t^2-t+1. Now how can i proceed?
That is a very good start. I suggest that you continue by completing the square, to get $(t-\tfrac12)^2 + \tfrac34.$ You know that t must lie between 0 and 1, so from that you can work out the range of the function.

3. ## Re: range of trigonometric function

Originally Posted by Opalg
That is a very good start. I suggest that you continue by completing the square, to get $(t-\tfrac12)^2 + \tfrac34.$ You know that t must lie between 0 and 1, so from that you can work out the range of the function.
Sorry! I have done so, but not understanding whether i should put value of t here. If i do so answer is 1 in both cases. Help

4. ## Re: range of trigonometric function

Ok, do you know the range of this function?

$(t-\tfrac12)^2 + \tfrac34.$

5. ## Re: range of trigonometric function

Originally Posted by pickslides
Ok, do you know the range of this function?

$(t-\tfrac12)^2 + \tfrac34.$
no!

6. ## Re: range of trigonometric function

do you know the range of t?

7. ## Re: range of trigonometric function

Well I would say the range of t is (0.75, $\infty$)

And as Opalg already suggested the range between 0 and 1 then the original functions range is the intersection of the 2.

I.e $\displaystyle [0,1] \cap \left[\frac{3}{4}, \infty\right) = \dots$

8. ## Re: range of trigonometric function

Originally Posted by ohm
Originally Posted by Opalg
That is a very good start. I suggest that you continue by completing the square, to get $(t-\tfrac12)^2 + \tfrac34.$ You know that t must lie between 0 and 1, so from that you can work out the range of the function.
Sorry! I have done so, but not understanding whether i should put value of t here. If i do so answer is 1 in both cases. Help
You have to think about what happens as t goes from 0 to 1. As t goes from 0 to 1/2, $(t-\tfrac12)^2$ goes from 1/4 down to 0. And as t goes from 1/2 to 1, $(t-\tfrac12)^2$ goes from 0 back up to 1/4. So the range of $(t-\tfrac12)^2$ will be the interval [0,1/4]. It follows that the range of $(t-\tfrac12)^2 + \tfrac34$ is the interval [3/4,1].