Having trouble on trig question

1. Oct 27, 2008

david18

Hi, I get stuck on the 3rd part of this question:

(i) Draw sketches of the functions sin x and $sin^2x$ over the range -360<x<360.
(ii) Explain why, for the range 0 < x < 90=2, sin x is smaller than tan x.
(iii) Using the equality $cos^2x=\frac{1}{2}(1+cos2x)$ or otherwise, express
$cos^4x$in terms of cos2x and cos4x.

My attempt at part iii:

I squared the left hand side of $cos^2x=\frac{1}{2}(1+cos2x)$ to get $cos^4x$ and therefore squared the right hand side as well, leaving the right hand side as $\frac{1}{4}(1+cos2x)^2$

I'm presuming I have to square the right hand bracket out but I'm unsure on what (cos2x)^2 becomes.

Any help would be appreciated

2. Relevant equations

3. The attempt at a solution

2. Oct 27, 2008

HallsofIvy

Staff Emeritus
If you know (because it was just given!) that
$$cos^2(a)= \frac{1}{2}(1+ cos(2a)$$
for any a, then what is $cos^2(2x)$?

3. Oct 27, 2008

david18

Thanks for the reply, following your method I ended up with $\frac{3}{8}+\frac{1}{2}cos2x+\frac{1}{4}cos4x$

Just out of interest, if I were to square cos2x would I have to use the knowledge given in the question or does it equal something obvious?

4. Oct 27, 2008

HallsofIvy

Staff Emeritus
Only the "obvious" $cos^4(2x)$! But then you could use the same identities to reduce that. It's just easier to use the already factored form.