# Trigonometry question find range

1. Sep 9, 2008

### fantasy

guys , following is the problem

(sinx)^4 + (cosx)^4

and

(sinx)^6 + (cosx)^6

they are asking for the range for each of them

i really have know clue how to solve this...any sugeestion please?....thanks for reading this!!

2. Sep 9, 2008

### snipez90

Both of them can be solved in the same way I think. Here is a hint on how to transform the first one.

$$a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2$$

3. Sep 9, 2008

### fantasy

so the other one will be

(a+b)^6 =(a^2 +b^2)^3- 3a^2b - 3ab^2

rite?..

i tried to solve and factorize the first one but cant

it becomes 2(sin^2(x))^2 - sin^2(x) + 1 ??

Last edited: Sep 9, 2008
4. Sep 9, 2008

### praharmitra

Use the property $$sin(2x) = 2sin(x)cos(x)$$

5. Sep 9, 2008

### fantasy

thanks praharmita... but there is not sin 2x ?

its ((sinx)^2)^2....

6. Sep 9, 2008

### praharmitra

you dont have to write everything in terms of sin(x) as you did. Use the original form as given by snipez90, where there is a $$2a^2b^2$$ term, and then use the identity of sin(2x)

7. Sep 10, 2008

### fantasy

ooo i see...thanks again....

is the ans 90<x<270 for the first one ?

8. Sep 10, 2008

### snipez90

I assume you mean degrees, in which case you are thinking about the domain but we want to focus on the range (the outputs of our function).

We have

$$sin^4(x) + cos^4(x) = (sin^2(x) + cos^2(x))^2 - 2sin^2(x)cos^2(x) = 1 - [2sin(x)cos(x)]sin(x)cos(x) = 1 - sin(2x)(\frac{1}{2}sin(2x)) = 1 - \frac{1}{2}sin^2(2x)$$

Now using what you know about the range of sin and the nature of squared quantities, determine the range of the final expression.