[Daven]

Got this for some revision for the upcoming exam, but it just didnt seem to fit anyhting i could work out.

Cos(A+40)Cos(A-10) = 0.5

And after some really bad maths, decided the answers would be somewhere to seek inspiration..to find that they get it right and dont actually help to much.
If someone could kindly show me the method i would be very grateful.
(0 < A <360 btw)

[Cexy]

cos(θ+Ø) = cosθcosØ - sinθsinØ
cos(θ-Ø) = cosθcosØ + sinθsinØ

Adding these together and dividing by two,

cosθcosØ = ½ [cos(θ+Ø) + cos(θ-Ø)]

Letting θ=A+40, and Ø=A-10, we get,

cos(A+40)cos(A-10) = ½ [cos(2A+30) + cos(50)] = ½

cos(2A+30) = 1 - cos(50)

2A + 30 = cos^-1(1-cos50)

A = ½ [cos^-1(1-cos50) - 30]

[Daven]

Nicely done Thankyou, i'll try to remember the way you did it..dont think our teacher really helped on this

[Mr. Pink]

its really important to remember the identities (in this case factor formulae)
that cexy used, they come into play later in maths (c4 especially)

[Daven]

Hmm got another question which i am having trouble with..(Im not that lame at maths, honest)

Find the equation of the tangent to the curve y=e ^ x/3 divided by x at the point 3. 1/3e

I've tried to differentiate the equation using quotient rule to get

dy/dx = 1/3x(e^x/3) - e^x/3 divided by x^2 (wow looks a lot more confusing here than on my paper)

I've then substituded x=3 into it to get 0/9 :| something seem wrong there. Could you help again please, point me to how the differentiation went wrong if its that or something that ive missed.
Once with the gradient use y-y1=m(x-x1) to get the equation (i think thats the right formula..something like that)

[Dimez]

dy/dx should be [1/3(e^(x/3) - e^(x/3)]x^2.

[Daven]

How so? if quotient rule is dy/dx = (v*du/dx - u*dv/dx) / v^2 then i dont see how you got to that

and 1/3e^x/3 - e^x/3 makes a negative, which is then made bigger by the x^2, and it didnt seem to fit with the way the book went silly maths

[Malik]

Originally Posted by Daven

How so? if quotient rule is dy/dx = (v*du/dx - u*dv/dx) / v^2 then i dont see how you got to that

and 1/3e^x/3 - e^x/3 makes a negative, which is then made bigger by the x^2, and it didnt seem to fit with the way the book went silly maths

I get

y=e^x/3/x
y=e^x/3x^-1

dy/dx= -e^x/3x^-2+ x^-1(1/3)e^x/3
e^x/3(-x^-2+x^-1/3)
e^x/3(1/3x-1/x²)

[Daven]

then you'd sub x=3 in to get the gradient? which from your equation comes out to be around 2.4 from (e^1 * (1-1/9))

then us y-1/3e = 2.4(x-3) which gives an equation of y = 2.4x-6.28 book says y=1/3x...this isnt going well

[Dimez]

Originally Posted by Daven

How so? if quotient rule is dy/dx = (v*du/dx - u*dv/dx) / v^2 then i dont see how you got to that

and 1/3e^x/3 - e^x/3 makes a negative, which is then made bigger by the x^2, and it didnt seem to fit with the way the book went silly maths

Oh sorry, that was a mistake on my part. I forgot to mutiply by the denominator.

[Daven]

Its not a problem, i dont mean to be annoying, but to me it seems like i've done everything right, but just cant get the answer, so will pick things apart and question them and as of yet i stil havent got it perfectly right

[Dimez]

dy/dx= [e^(x/3) ( x/3 - 1)]x^2

sub in x=3 to calculate the gradient===> gradient is 0 because e(1-1)= 0

y - e/3=0(x - 3)

y= e/3 which dosen't seem to correspond with the given answer.

[Daven]

given answer is y=1/3e So by the looks of that you've done it....so there is no gradient..and y is a straight line, therefore the fact that you get dy/dx = 0 just means its right and you have to carry on. Nicely done

[Malik]

Originally Posted by Daven

then you'd sub x=3 in to get the gradient? which from your equation comes out to be around 2.4 from (e^1 * (1-1/9))

then us y-1/3e = 2.4(x-3) which gives an equation of y = 2.4x-6.28 book says y=1/3x...this isnt going well

e¹(1/9-1/9)=0
.:. gradient=0

I dont know how you got 2.4

[Daven]

I get

y=ex/3/x
y=ex/3x-1

dy/dx= -ex/3x-2+ x-1(1/3)ex/3
ex/3(-x-2+x-1/3)
ex/3(1/3x-1/x²)

It was the last bit line (which i've actually just read correctly) I guess i read it has 1/3 of X, making 1, then - 1/9 and so on. Sorry for that error