# Help on P2 Trig QsWatch

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#1
1) Express 2cosx + 5sinx in the form Rcos(x - alpha), where R>0 and 0 < alpha < 90. Give the values of R and alpha to 3 significant figures.

2) Given that in the expansion (1+x)^n, where n>2, the coefficient of x is 8 and of x^2 is 30. Calculate the value of n and the value of a.

Any help will be appreciated. Thank you! Love Pinkfairy xxx
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15 years ago
#2
Express 2cosx + 5sinx in the form Rcos(x - alpha), where R>0 and 0 < alpha < 90. Give the values of R and alpha to 3 significant figures

Rcosalpha.cosx + Rsinalpah.sinx = Rcos(x-alpha)

Rcosalpha = 2 (call eqtn1)
Rsinalpha = 5(call eqtn2)

eqtn2/eqtn1 gives tan alpha = 5/2 -> alpha=68.2

eqtn1^2 + eqtn2^2 = R^2 = 50
R=sqrt50
R=7.07

Hope they're right! Soz, have to go now, will answer your second quetsion later if I have time
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#3
(Original post by pinkfairy)
1) Express 2cosx + 5sinx in the form Rcos(x - alpha), where R>0 and 0 < alpha < 90. Give the values of R and alpha to 3 significant figures.
I have just worked out the second Q but I'm not sure about the first. I have got 5.39cos(x - 68.2) as my answer, will someone tell me if that's correct or not? Love Pinkfairy xxx
0
15 years ago
#4
(Original post by pinkfairy)
1) Express 2cosx + 5sinx in the form Rcos(x - alpha), where R>0 and 0 < alpha < 90. Give the values of R and alpha to 3 significant figures.

2) Given that in the expansion (1+x)^n, where n>2, the coefficient of x is 8 and of x^2 is 30. Calculate the value of n and the value of a.

Any help will be appreciated. Thank you! Love Pinkfairy xxx
Rcos(x-a) = Rcosxcosa + Rsinxsina

Therefore

Rcosa = 2
Rsina= 5

tana = 5/2 a = 68.2 degrees

R = Root(5^2 + 2^2)
= root 29

(1+x)^n = 1+xn + (n/2)x^2

n=8

There is no a in that expansion??
0
15 years ago
#5
(Original post by imasillynarb)
Rcos(x-a) = Rcosxcosa + Rsinxsina

Therefore

Rcosa = 2
Rsina= 5

tana = 5/2 a = 68.2 degrees

R = Root(5^2 + 2^2)
= root 29

(1+x)^n = 1+xn + (n/2)x^2

n=8

There is no a in that expansion??
It should be (1 + ax)^n, IIRC. After you do the expansion, all you have to do is compare coefficients of x and x^2 and then solve those simultaneous equations.
0
15 years ago
#6
(Original post by Nylex)
It should be (1 + ax)^n, IIRC.
Then:

(1+ax)^n = 1 + anx + (n/2) (a^2.x^2)

an = 8 a=8/n

a^2. n(n-1)/2 = 30

2a^2. n^2 - n = 60

= 64/n^2(n^2 - n) = 60

64 - 64/n = 60

64n - 64 = 60n
4n = 64
n = 16
a=1/2

Think thats right
0
15 years ago
#7

(1+an)^n

Expanded, this becomes,

1+nax + n(n-1)(ax)²/2 + ...

Then,

na=8
½n(n-1)a²=30

using a=8/n and substituting,

½n(n-1)64/n²=30
32(n-1)=30n
32n-30n=32
2n=32
n=16
a=½
0
15 years ago
#8
I got the right answer but I did something wrong
0
#9
(Original post by Fermat)

(1+an)^n

Expanded, this becomes,

1+nax + n(n-1)(ax)²/2 + ...

Then,

na=8
½n(n-1)a²=30

using a=8/n and substituting,

½n(n-1)64/n²=30
32(n-1)=30n
32n-30n=32
2n=32
n=16
a=½
I have recently take up maths again hence the difficulty but thanks so much for your help Love Pinkfairy xxx
0
15 years ago
#10
(Original post by pinkfairy)
1) Express 2cosx + 5sinx in the form Rcos(x - alpha), where R>0 and 0 < alpha < 90. Give the values of R and alpha to 3 significant figures.

2) Given that in the expansion (1+x)^n, where n>2, the coefficient of x is 8 and of x^2 is 30. Calculate the value of n and the value of a.

Any help will be appreciated. Thank you! Love Pinkfairy xxx
how bizarre.

i just did that first question (but in radians).
0
15 years ago
#11
(Original post by imasillynarb)
I got the right answer but I did something wrong
Lol.
0
15 years ago
#12
Ignore the fact that I said R=sqrt50! Dunno where that came from! grrrrr. I was in the shower just after I'd post that and I suddenly thought "Duhh why did I do that it's definitely sqrt29!". Grrr i hate the fact that sometimes I rush things and make stupid mistakes!
0
15 years ago
#13
(Original post by Hoofbeat)
Grrr i hate the fact that sometimes I rush things and make stupid mistakes!
That's always my problem... I make stupid mistakes and can't spot that there's anything wrong with them until I'm told it's wrong... then I spot it straight away
0
15 years ago
#14
(Original post by Pixelfairy #1)
That's always my problem... I make stupid mistakes and can't spot that there's anything wrong with them until I'm told it's wrong... then I spot it straight away
Don't we all!
That's just happened to me!
0
#15
Anyone help on this one?:

A root of the equation e^x-7x=0 lies in the interval N<x<N+1 where N is an integer. Find 2 values for N.

0
15 years ago
#16
(Original post by pinkfairy)
Anyone help on this one?:

A root of the equation e^x-7x=0 lies in the interval N<x<N+1 where N is an integer. Find 2 values for N.

I had one like this before the holidays.

I think what you have to do is get 2 iteration equations out of it, and then N..N+1 is the range which the roots are in once you put in a number and see where its going to.

Hope that helps
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#17
(Original post by Pixelfairy #1)
I had one like this before the holidays.

I think what you have to do is get 2 iteration equations out of it, and then N..N+1 is the range which the roots are in once you put in a number and see where its going to.

Hope that helps
I thought maybe iteration would solve it, but... Perhaps I can log both sides? Thank you for you help Love Pinkfairy xxx
0
15 years ago
#18
You need to sketch a graph of e^x and 7x and find the intersection points.
0
15 years ago
#19
(Original post by pinkfairy)
Anyone help on this one?:

A root of the equation e^x-7x=0 lies in the interval N<x<N+1 where N is an integer. Find 2 values for N.

Seeing as it's only p2, this won't really help, but if you use the newton-raphson method, and starting values of 1 and 3, you find that the roots are about 3.066 and 1.691, so N can be 1 and 3. I can't remember how they'd expect you do find this in p2 though!
0
15 years ago
#20
(Original post by Bezza)
Seeing as it's only p2, this won't really help, but if you use the newton-raphson method, and starting values of 1 and 3, you find that the roots are about 3.066 and 1.691, so N can be 1 and 3. I can't remember how they'd expect you do find this in p2 though!
I think you got the decimal point in the wrong place! 0.1691 not 1.691.
0
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