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1. Hello,

The question is as follows.

Solve 2cosx + 5sinx=4 by expressing in the form of R cos(2x - (alpha)).

I try expanding the expression but still cannot do it.

Hello,

The question is as follows.

Solve 2cosx + 5sinx=4 by expressing in the form of R cos(2x - (alpha)).

I try expanding the expression but still cannot do it.

Are you sure it says R cos(2x - a) rather than R cos(x - a)?
Hello,

The question is as follows.

Solve 2cosx + 5sinx=4 by expressing in the form of R cos(2x - (alpha)).

I try expanding the expression but still cannot do it.

I think this has been discussed before
https://www.thestudentroom.co.uk/sho....php?t=2741342

Some discussion about whether its a "good question" but if you want to use a double angle formula, you need to get some cos^2, sin^2 type terms, so square both sides of the expression and start from there ... Doesn't sound like its a particularly fruitful journey :-)
4. (Original post by mqb2766)
I think this has been discussed before
https://www.thestudentroom.co.uk/sho....php?t=2741342

Some discussion about whether its a "good question" but if you want to use a double angle formula, you need to get some cos^2, sin^2 type terms, so square both sides of the expression and start from there ... Doesn't sound like its a particularly fruitful journey :-)
Well, I did not find that post when using google,sorry for that.
And also it is the same question in the same book(my version is the latest).
As the question does not change after 3years,I think that maybe this is not a typo somehow..
The question also does not mention what should be expressed in the form of r cos(2x+a) .
In this section only this question bothers me..
Well, I did not find that post when using google,sorry for that.
And also it is the same question in the same book(my version is the latest).
As the question does not change after 3years,I think that maybe this is not a typo somehow..
The question also does not mention what should be expressed in the form of r cos(2x+a) .
In this section only this question bothers me..
No problem about not finding it, I was simply pointing out that there was some discussion a couple of years ago.

Did you try squaring the expressions in order to use the double angle formulas? If so, what problems did you find? I've not derived it yet, but that would seem to be the obvious way to start if you've got to use a double angle formula.
6. this is a typical "harmonic form" or "wave form" problem.

in general if you have f(x) = acosx + bsinx

you can write it as f(x) = Rcos( x - α )

where R = √ ( a2 + b2 )

and tan α = b/a
7. (Original post by mqb2766)
I think this has been discussed before
https://www.thestudentroom.co.uk/sho....php?t=2741342

Some discussion about whether its a "good question" but if you want to use a double angle formula, you need to get some cos^2, sin^2 type terms, so square both sides of the expression and start from there ... Doesn't sound like its a particularly fruitful journey :-)
And also I try to multiply cos x to the entire equation to form:
2 cos^2 x + 5 sin x cos x - 4 cos x = 0

By comparing coefficients,
2r cos a =2
2r sin a =-5
r cos a =4 cos x

But I still cannot get the correct answer.
Maybe I will try squaring and see if this works
8. (Original post by the bear)
this is a typical "harmonic form" or "wave form" problem.

in general if you have f(x) = acosx + bsinx

you can write it as f(x) = Rcos( x - α )

where R = √ ( a2 + b2 )

and tan α = b/a
The question asked for r cos (2x+a) not the usual r cos (x+a)
The question asked for r cos (2x+a) not the usual r cos (x+a)
could you post the problem as it appeared in the book please ?
10. (Original post by mqb2766)
No problem about not finding it, I was simply pointing out that there was some discussion a couple of years ago.

Did you try squaring the expressions in order to use the double angle formulas? If so, what problems did you find? I've not derived it yet, but that would seem to be the obvious way to start if you've got to use a double angle formula.
In the post three years ago somebody gives an answer by squaring,but as somebody pointed out ,the person did not expand completely,
so it should be:
r cos a =-25
r sin a =-10
2r sin a =-21
But it is inconsistent.
So how?
11. (Original post by the bear)
could you post the problem as it appeared in the book please ?
The question is the same question that has been posted 3 years ago.So you can browse that thread(https://www.thestudentroom.co.uk/sho....php?t=2741342 )The asker had post the image of the question from the book.My book is same as the asker but just a newer version and the question did not change at all.
In the post three years ago somebody gives an answer by squaring,but as somebody pointed out ,the person did not expand completely,
so it should be:
r cos a =-25
r sin a =-10
2r sin a =-21
But it is inconsistent.
So how?
It doesn't work, and the book has a typo.
13. (Original post by Prasiortle)
It doesn't work, and the book has a typo.
But why after 3 years in the latest version the question is still the same after updating?
Well,actually this section my teacher still does not teach so maybe she will say about it after that.
But why after 3 years in the latest version the question is still the same after updating?
Well,actually this section my teacher still does not teach so maybe she will say about it after that.
The Edexcel books are terrible and errors often don't get corrected.
15. (Original post by Prasiortle)
The Edexcel books are terrible and errors often don't get corrected.
For your information,that is not a Edexcel book.The publisher is Pelangi(I bet you never heard before)
And also I try to multiply cos x to the entire equation to form:
2 cos^2 x + 5 sin x cos x - 4 cos x = 0

By comparing coefficients,
2r cos a =2
2r sin a =-5
r cos a =4 cos x

But I still cannot get the correct answer.
Maybe I will try squaring and see if this works
Maybe I can let r cos (2x+a) become r cos (x+(x+a)), but seems it will return to r cos (x+a) in the end.
Maybe I can let r cos (2x+a) become r cos (x+(x+a)), but seems it will return to r cos (x+a) in the end.
Yes, it's not going to help you. The fact is pretty much everyone who has looked at this has concluded that the problem is wrong.
18. Let's have a try. For brevity,
c = cos(x)
s = sin(x)
c2 = cos(2x)
s2 = sin(2x)

Square the original expression
2*c + 5*s = 4
becomes
4*c^2 + 25*s^2 + 20*c*s = 16

Using the standard double angle formula
s2 = 2*s*c
c2 = 1 - 2*s^2 = 2*c^2 - 1

This becomes
4/2*(c2 + 1) + 25/2*(1 - c2) + 20/2*s2 = 16
or
-10.5*c2 + 10*s2 = 1.5

I think
r*cos(2x - alpha) = r*cos(alpha)*cos(2x) + r*sin(alpha)*sin(2x)
r = 14.5

and solve gives x = 1.923 rad which is correct subbing back in the original expression.
19. (Original post by mqb2766)
Let's have a try. For brevity,
c = cos(x)
s = sin(x)
c2 = cos(2x)
s2 = sin(2x)

Square the original expression
2*c + 5*s = 4
becomes
4*c^2 + 25*s^2 + 20*c*s = 16

Using the standard double angle formula
s2 = 2*s*c
c2 = 1 - 2*s^2 = 2*c^2 - 1

This becomes
4/2*(c2 + 1) + 25/2*(1 - c2) + 20/2*s2 = 16
or
-10.5*c2 + 10*s2 = 1.5

I think
r*cos(2x - alpha) = r*cos(alpha)*cos(2x) + r*sin(alpha)*sin(2x)
r = 14.5

and solve gives x = 1.923 rad which is correct subbing back in the original expression.
Well,just a little problem,the question wants r cos(2x+a) not r cos (2x-a)
Well,just a little problem,the question wants r cos(2x+a) not r cos (2x-a)
Argh :-) I'll let you figure that one out.

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