Having trouble on trig question

  • Thread starter david18
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  • #1
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Hi, I get stuck on the 3rd part of this question:

(i) Draw sketches of the functions sin x and [itex]sin^2x[/itex] 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 [itex]cos^2x=\frac{1}{2}(1+cos2x)[/itex] or otherwise, express
[itex]cos^4x[/itex]in terms of cos2x and cos4x.

My attempt at part iii:

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

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



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The Attempt at a Solution

 

Answers and Replies

  • #2
HallsofIvy
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If you know (because it was just given!) that
[tex]cos^2(a)= \frac{1}{2}(1+ cos(2a)[/tex]
for any a, then what is [itex]cos^2(2x)[/itex]?
 
  • #3
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Thanks for the reply, following your method I ended up with [itex]\frac{3}{8}+\frac{1}{2}cos2x+\frac{1}{4}cos4x[/itex]

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
HallsofIvy
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Only the "obvious" [itex]cos^4(2x)[/itex]! But then you could use the same identities to reduce that. It's just easier to use the already factored form.
 

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