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# P3 Question watch

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1. Q1:
Ok L1 has directional vector -4i+2j+7k
and L2 has directional vector i-2j-2k

Calculate the "acute" angle between L1 and L2.

I managed it using dot product, but ended up with a non-acute angle, yet it requires an acute angle . Please help.

Q2:
Also, integrate: cos3xcosx.dx ---> this shouldnt be a problem I split up cos3x in terms of cos2x and cosx just wanted to check my answer.
2. For q1, just do 180 - answer to get acute angle. You got an obtuse becuase your lines were not both pointing towards the angle or away from the angle.

If the questions are hard they will state that the angle is acute. If it is an easy question they will expect you to have the vectors pointing the right way
3. (Original post by ResidentEvil)
Q1:
Ok L1 has directional vector -4i+2j+7k
and L2 has directional vector i-2j-2k

Calculate the "acute" angle between L1 and L2.

I managed it using dot product, but ended up with a non-acute angle, yet it requires an acute angle . Please help.

Q2:
Also, integrate: cos3xcosx.dx ---> this shouldnt be a problem I split up cos3x in terms of cos2x and cosx just wanted to check my answer.
Acute angle is 180-angle you got

3x = A + B/2
6x = a + b

A-B /2 = x
2x = a-b
a=2x+b
6x = 2x + b + b
b = 2x
a = 4x

Integral of (cos4x + cos2x) / 2
=1/8sin4x + 1/4sin2x + C
4. Not sure about part 2, havent got paper but I would do:

cos3xcosx = cos(2x+x)cosx
= (cos2xcosx - sin2xsinx)cosx
= [(cos^2x-sin^2x)cosx - (sin2xcosx - cos2xsinx)]cosx
= [(2cos^2x - 1)cosx - (2sinxcos^2x - {cos^2x-sin^2x}sinx]cosx
...

is this what you did???
5. (Original post by imasillynarb)
Acute angle is 180-angle you got

3x = A + B/2
6x = a + b

A-B /2 = x
2x = a-b
a=2x+b
6x = 2x + b + b
b = 2x
a = 4x

Integral of (cos4x + cos2x) / 2
=1/8sin4x + 1/4sin2x + C
What is all this A and B stuff??? where did it come from???

Is it some of those identities in the formula book

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