Maths edexcel a level trig chapter 7

Could someone explain to me how to solve this question. I understand I need to rearrange both terms to just cos terms but am stuck.

https://imgur.com/a/31FcLfI

Could you also explain how the solution bank got the answers below for cos squared and sin squared

https://imgur.com/a/hD2QHEl

Reply 1

mqb2766

Original post

by saminasb3

Could someone explain to me how to solve this question. I understand I need to rearrange both terms to just cos terms but am stuck.
https://imgur.com/a/31FcLfI
Could you also explain how the solution bank got the answers below for cos squared and sin squared
https://imgur.com/a/hD2QHEl

It helps to have a rough idea of what trig identity works so here you have soemthing like
cos^2(x)
and
sin^2(x)
and both of those map to something like cos(2x). So raising the trig term to a power is similar to multiplying the angle by the same value. So here x = .... and just apply the relevant identities.

The soln bank uses the half angle identities (which are standard), or you could think about them in terms of double angles to get the same.

Reply 2

username79352

When you say that raising the trig term to a power does that mean that cos^2x is the same as 2cosx?

Reply 3

mqb2766

Original post

by saminasb3

When you say that raising the trig term to a power does that mean that cos^2x is the same as 2cosx?

Certainly not as 2(cos(x)) and (cos(x))^2 are obviously different, as you could simply sub cos(x)=1/2 and one is 1 and the other 1/4.

In a sense you seem to not undertand the model solution as they use the half angle identities to (roughly) transform cos^2(x/2) to cos(x). You must be able to work the question through with the relevant identities given (its just a substitution) and basic algebra?

The hint was more about attempting the question without the model solution and having a hint about how to start / does the question make sense / which identities to use. So the hint was
sin^2(x) ~ cos(2x)
cos^2(x) ~ cos(2x)
so when you square a trig value (or cube or ...) its roughly equivalent to the trig of a double angle (or triple or ..). The "roughly" can hide important details but here we have an expression which involves the square of half angles and we want to do an identity transformation to a "full" (or double the half angle) angle so
sin^2(x/2) ~ cos(x)
cos^2(x/2) ~ cos(x)
Using the hint, you should simply recognise that its a standard half angle identity and look it up then a couple of lines of algebra using the correct identities, which are in the formula sheet.

So you have the usual double angle identities
cos(2x) = cos^2(x)-sin^2(x) = 1 - 2sin^2(x) = 2cos^2(x)-1
sin(2x) = 2sin(x)cos(x)
so sin or cos of a double angle, 2x, corresponds to roughly the product of two trig terms of x. The "roughly" should be clear from the example in that there may be a constant, a multiplier on the trig term or the product of two trig terms instead of squaring.

Reply 4

username79352

https://imgur.com/a/VIxhO2Y

I got confused by the sin and cos and try to convert it into cos2x that way. So for the double angle formulae it can only be in cos form not cos^2 when you’re writing as vos2x ie you can’t write it as cos^2(2x) and the angle is always halved in the expansion? Or double if going backwards.

Reply 5

mqb2766

Original post

by saminasb3

https://imgur.com/a/VIxhO2Y
I got confused by the sin and cos and try to convert it into cos2x that way. So for the double angle formulae it can only be in cos form not cos^2 when you’re writing as vos2x ie you can’t write it as cos^2(2x) and the angle is always halved in the expansion? Or double if going backwards.

Firstly, it looks like your working is correct (2nd part of your answer),

Im not too sure what youre trying to ask (first part) but you have the double angle and half angle identities
https://www.onlinemathlearning.com/half-angle-identities.html
the first two half angle identities are pretty much the ones given in the model solution though theyve square both sides. So the identities are
trig (half angle) = f (trig(full angle))
for some simple function f(). Similarly the double angle identities are written as
trig (double angle) = f(trig(full angle))

However as an example, consider the second double angle identity so
cos(2x) = 1 - 2sin^2(x)
this is true for any angle x so I can sub in x/2 to get
cos(x) = 1 - 2sin^2(x/2)
and rearrnage to get
sin^2(x/2) = (1 - cos(x))/2
so the usual half angle identity. So the double and half angle identities are basicallly they same, its just theyre written with a different subject on the left hand side.

For your example of what cos^2(2x) might go to, then (without using the pythagoran identitiy which can deflate a power by 2) youd expect to be able to express it as somehting like cos(4x) or cos^4(x) and appropriate lower powers/multiple angles Here the cos could be sin or a combination of the two or it involve a multiplier or consntant or ... but having a slightly bigger picture can help about how the powers and angles are related in trig expressions. But you have to use the proper identities to grind through the actual algebra.

Reply 6

username79352

Firstly thank you so much for your help.

For the half angle formulae can I just sub them in after stating them or am I expected to first find them by rearranging and then sub them in?

And I meant that if a angle is in the form cos^2 x like in this question I should think about the cos2x rule and rearranging rather than actually applying the power if they just want the answer in a simpler form

Reply 7

mqb2766

Original post

by saminasb3

Firstly thank you so much for your help.
For the half angle formulae can I just sub them in after stating them or am I expected to first find them by rearranging and then sub them in?
And I meant that if a angle is in the form cos^2 x like in this question I should think about the cos2x rule and rearranging rather than actually applying the power if they just want the answer in a simpler form

I thought the double / half angle were in the formula sheet but from a quick check theyre not (edexcel), but somewhat perversely the hyperbolic double angles are. So check your formula sheet but if not, the double angles are a 1-liner from the usual addition identites, which are in the formula sheet, and then the half angles are a simple rearrangement (above).

In the original reply I was trying to get you to note that
cos^2(x) <-> cos(2x)
so you can go either way and the <-> is approximate. So for this question, youre asked to put cos^2(x/2) => cos(x)
and your first reaction should be "of course I can and theres an easy double/half angle identity" because
cos^2(x/2) => cos(2*x/2) => cos(x)
where in the above => is approximate. Then its just a case of grinding cos^2(x/2) formula and subbing it in. You could also grind sin^2(x/2) or put the original expression in terms of just cos^2(x/2) using the pythagoras identity which is probably slightly simpler

I think about trig functions sin and cos as basically exponentlals, as pretty much by defn you have
exp^2(x/2) = exp(x)
its little different for sin and cos trig functions (apart from its a bit more approximate).

Reply 8

username79352

Ohh ok thank you so much for all your help! :smile:

Reply 9

mqb2766

Original post

by saminasb3

Ohh ok thank you so much for all your help! :smile:

Note that it might have been slightly simpler to write what we want to show as
2cos^2(x) - 4sin^2(x) => cos(2x)
then its more obvious that a) is almost a 1 liner as
2cos^2(x) - 4sin^2(x) = 3(cos^2(x) - sin^2(x)) - (cos^2(x)+sin^2(x)) = 3cos(2x) - 1
just using the usual pythagorean identity and the "basic" double angle cos(2x) = cos^2(x)-sin^2(x). The latter is just the addition identity. Its a fair bit simpler than explicitly using the half angle identities, esp if you have to derive them.

Got sidetracked by the half angle identities in the model solution.