Double Angle Identities (2 Viewers)

[Arucerious]

I have an upcoming 3U test on trig and I'm not too sure which of the 3 cos2x expansions to use and when.
e.g cos2x = cos^2x - sin^2x, 1-2sin^2x, 2cos^2x - 1

Can someone explain in what situation would I use one expansion over the other?

Example question: 1+cos(180degree + 2x)
How come to simplify this question I was told to use the (1-2sin^x) expansion but not the others?

 

[InteGrand]

Example question: 1+cos(180degree + 2x)

You can just simplify that by using that (this can be seen by translating the graph of , just observing the graph, or using an "ASTC" diagram.)

 

[PhysicsMaths]

Well let's see
1+cos(180+2x)
= 1+cos180cos2x-sin180sin2x
= 1-cos2x
= 1-(1-2sin^2x)
= 2sin^2x

The reason why you used this variant was to remove the 1

 

[Crisium]

Use the form based on what the question is asking

1) If it is a prove question then look at the right hand side which will usually give you a hint as to what version of it you will have to apply.

2) If it is a simplification question then use the form that will be most convenient when expanding brackets and simplifying.

3) When integrating something like sin^2x, you're going to have to change it to a form where it can be integrated and so you will have to rearrange the identity cos2x = 1 - 2sin^2x into 0.5(1 - cos2x) and sub it in.

 

[Crisium]

Well let's see
1+cos(180+2x)
= 1+cos180cos2x-sin180sin2x
= 1-cos2x
= 1-(1-2sin^2x)
= 2sin^2x

The reason why you used this variant was to remove the 1

Alternatively as integrand state cos(180 + x) = -cosx <-- This is because the cosine function is negative in the third quadrant

1 + cos(180 + 2x)

= 1 + (-cos2x) < --- Using what I stated above

= 1 - cos2x

= 1 - (1 - 2sin^2x) OR = 1 - (2cos^2x - 1) OR = 1 - (cos^2x - sin^2x)

= 1 - 1+ 2sin^2x = 1 - 2cos^2x + 1 = 1 - cos^2x + sin^2x

= 2sin^2x = 2-2cos^2x = sin^2x + sin^2x

= 2(1 - cos^2x) = 2sin^2x

= 2sin^2x

There are heaps of other ways to do these questions but its best to use the one that will give the you the simplest answer

 

[InteGrand]

Alternatively as integrand state cos(180 + x) = -cosx <-- This is because the cosine function is negative in the third quadrant

1 + cos(180 + 2x)

= 1 + (-cos2x) < --- Using what I stated above

= 1 - cos2x

= 1 - (1 - 2sin^2x) OR = 1 - (2cos^2x + 1) OR = 1 - (cos^2x - sin^2x)

= 1 - 1+ 2sin^2x = 1 - 2cos^2x - 1 = 1 - cos^2x + sin^2x

= 2sin^2x = -2cos^2x = sin^2x + sin^2x

= 2sin^2x

There are heaps of other ways to do these questions but its best to use the one that will give the you the simplest answer

(It should be 2cos² x – 1 for the middle set, not 2cos² x + 1.)

 

[Crisium]

(It should be 2cos² x – 1 for the middle set, not 2cos² x + 1.)

My apologies

As you can see I haven't been getting much sleep :/

 

[laters]

A general guide:

You can rearrange the equations to get:



Most of the manipulations required will contain 1+cos2x and 1-cos2x and the identities above are almost always the required versions.

For you question you could do two things:
(1) Use the double angle immediately:

(2) Use the relation cos(180+x)=-cos(x):