Spesh Help...

[tiff_tiff]

hello, can somebody please help me with 18b) thankyou

[e^1]

Part A helps part B

Part B, do it yourself first

[tiff_tiff]

thank you thank you

[tiff_tiff]

sin^2(x)cos^2(x) = 1
solve for x?

and lastly, how do you know that l(5/2 + 2cos(x))l = 5/2 + 2cos(x)
basically, the modulus of a number is the number?

thanks in advance

[brightsky]

1. notice that sin(x)cos(x) = 1/2 sin(2x). so sin^2(x)cos^2(x) = 1 becomes (1/2 sin(2x))^2 = 1. solve as per usual.

2. consider the equation mod(x) = x. imagine the graphs in your head. notice how the equation only holds for when x >= 0 (x greater or equal to 0). in general, the equation mod(blah) = blah only holds for when blah >= 0. so in your case, the equation gets reduced to the inequation 5/2 + 2cos(x) >= 0. solve as per usual.

[tiff_tiff]

yup yup i get it now thanks you so so much

[tiff_tiff]

im sorry but...

why is that with:
2sin(x)cos(x) = sin(x), if the question is to find x, why can't i just simply divide both sides by sin(x) to get 2cos(x)=1 and then solve from there?
I have to go 2sin(x)cos(x) - sin(x) = sin(x)[2cos(x)-1] and solve from there... (which i know how to do)

and:

As cos(x) = 2/3 , a . b = 2/3 b . b [this is related to vectors], can you please explain this statement?


Thank you

[b^3]

We avoid dividing by (or anything involving for that matter) as you can start losing solutions. In the case above if you were to divide by , you lose all the solutions associated with , so you need to rearrange, factorise and then use the null factor law instead.

We have the dot product, , for the above to work the magnitude of would have to be equal to the magnitude of . I feel there is more to the question, do you have more context or information for it? (it might be using the fact that .

[keltingmeith]

Building off of what b^3 has said, here's a simple exercise so you can see your loss of solutions:

Solve by dividing by , and then do it by taking to the other side, factorising and applying the null factor theorem.

Do the same thing for - divide by x one time, than solve by taking x to the other side another time.

You should see that when you divide you get only one of the solutions you'd get if you solve by the null factor theorem. I hope that gives some insight as to why we generally don't divide by variables.

[tiff_tiff]

thank you everyone

that cos thing is relate to 6cii)

[e^1]

One way to go through that question is to find OF and AE. If the dot product of these two vectors is 0, then you've shown it.

Work it out first

I mean it...

Last warning!

Some things to know before we show that OF is perpendicular to AE:



Working out:


Therefore OF is perpendicular to AE, since the dot product of the two is 0.

[Bestie]

Hello I was wondering:

what is tan^-1(2) + tan^-1(1/2)?     pi/2 is the ans?

and

tan (cos^-1(12/13) - sin^-1(4/5)) = -33/56 - how do i prove that?

thank you

[EspoirTron]

Hello I was wondering:

what is tan^-1(2) + tan^-1(1/2)?

pi/2 is the ans?

thank you

tan^-1(2) + tan^-1(1/2) = arctan(2) + arctan(1/2)

arctan(2) = pi/2 - arctan (1/2)

It thus follows that:

arctan(2) + arctan(1/2) = pi/2 - arctan(1/2) + arctan(1/2) = pi/2 , as the solution suggests.

[kinslayer]

tan (cos^-1(12/13) - sin^-1(4/5)) = -33/56 - how do i prove that?

thank you

If arccos(12/13) = t then cos t = 12/13 and sin t = 5/13 (draw a triangle)

Likewise if arcsin(4/5) = u then sin u = 4/5 and cos u = 3/5 (as above)

Now the solution wants you to find the value of tan(t - u) and you can use these to do it.

[brightsky]

Hint: Note that arccos(12/13) = arctan(5/12) and arcsin(4/5) = arctan(4/3). Now use compound angle formula for tangent.