- #1
david18
- 49
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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
(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