Specialist 1/2 Question Thread!

[PolySquared]

If

Then what is the value of 

[TheAspiringDoc]

If

Then what is the value of 

Just use the double angle formula twice. You should actually get 2 answers within the principal value of -Pi , pi

[PolySquared]

Just use the double angle formula twice. You should actually get 2 answers within the principal value of -Pi , pi

What do you mean by use it twice? Can you please demonstrate?

[RuiAce]

What do you mean by use it twice? Can you please demonstrate?

\begin{align*}\sin 4\theta &= 2\sin2\theta\cos2\theta\\ &= 4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)\end{align*}

[TheAspiringDoc]

Hmm rui's method is correct but a bit tricky IMO. Just let 2theta = x and then find the double angle of sin(2x).

[PolySquared]

Can I please get some help on how to answer this?

√3 cos(2x) − sin(2x) in the form r cos(2x + α).

[Sine]

Can I please get some help on how to answer this?

√3 cos(2x) − sin(2x) in the form r cos(2x + α).

√3 cos(2x) − sin(2x) =  r cos(2x + α).
 RHS =  r cos(2x + α).
          = rcos(2x)cos( α) - r sin(2x)sin( α)
 
LHS = √3 cos(2x) − sin(2x)

LHS = RHS
√3 cos(2x) − sin(2x) = rcos(2x)cos( α) - r sin(2x)sin( α)
 From here we can see that
√3 cos(2x)  = rcos(2x)cos( α)
Hence √3 = rcos( α) 
√3/r = cos( α)
  α = cos-1(√3/r)

Also we can see
sin(2x) = r sin(2x)sin( α)
1 = rsin(α)     
1/r = sin(α) 
α = sin-1(1/r)

Therefore,   sin-1(1/r) = α = cos-1(√3/r)

sin-1(1/r) = cos-1(√3/r)

From here you should be able to eyeball that r = 2 and also that  α = π/6
So now we have
√3 cos(2x) − sin(2x) =  rcos(2x + α)
√3 cos(2x) − sin(2x) =  2cos(2x + π/6).

[PolySquared]

Hi could I get some help on this question?

Write in the form

My answer is

but the textbook answer has 5pi/4 instead of pi/4.

[jazzycab]

Hi could I get some help on this question?

Write in the form

My answer is

but the textbook answer has 5pi/4 instead of pi/4.

Multiplying the entire expression by \(\frac{\sqrt{2}}{\sqrt{2}}\) we get:

If we then substitute for the exact values of \(\sin{\left(\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\) and \(\cos{\left(\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\), we get:

From here, we can recognise that what is in the brackets is simply the compound-angle formula for \(\sin{\left(a-b\right)}\):

But in this expression, our argument is not quite what we need (i.e. we need \(3x\) to be positive). We can use the fact that \(\sin{\left(-x\right)}=-\sin{\left(x\right)}\) to rectify this:

But now the sign of the entire expression is incorrect. Here we can use the fact that \(\cos{\left(-\pi\right)}=-1\) and \(\sin{\left(-\pi\right)}=0\) to force another compound angle formula for sine:


I'm certain that there is a more elegant way of doing this, but that's the way it fell out when I looked at it.

[Yiyiyi]

Four teachers decide to swap desks at work. How many ways can this be done if no teacher is to sit at their previous desk?

[PolySquared]

Hi guys can I please get some help on this question.

Let A = (5,1), B = (0,4) and C = (-1,0). Find:

D such that vector AB = vector CD

[RuiAce]

This gives \(d_2 = 3\) and \(d_1 = -6\), assuming I did not mess up basic arithmetic

[PolySquared]

This gives \(d_2 = 3\) and \(d_1 = -6\), assuming I did not mess up basic arithmetic

Thank you so much

[RuiAce]

Four teachers decide to swap desks at work. How many ways can this be done if no teacher is to sit at their previous desk?

Let teacher 1 originally occupy desk 1, teacher 2 originally occupy desk 2 and etc.
There are 3 ways this can happen, since we have 3 other teachers.

Now, without loss of generality, assume that teacher 2 occupies desk 1.
Case 1: Teacher 1 occupies desk 2.
Then, teacher 3 must occupy desk 4 and vice versa, so this case can only happen 1 way.

Case 2: Either one of teachers 3 or 4 occupy desk 2. Note that there are 2 ways this can happen, since we have 2 other teachers.
Then, without further loss of generality, assume that teacher 3 occupies desk 2.

Subcase 2.1: Teacher 1 occupies desk 3.
This can't happen, because then teacher 4 occupies their own desk.

Subcase 2.2: Teacher 4 occupies desk 3.
Then teacher 1 must occupy desk 4. Only 1 way this can happen.
\[ \therefore \text{Ans: }3\big(1 + 2(0+1) \big) = 9.\]
____________________________________________________________

The total outcomes is just the number of ways the teachers can be assigned to sit anywhere, which is just 4! = 24.

The inclusion-exclusion principle becomes necessary because we double count the number of ways more than 1 teacher can sit at their own seat.
# of ways at least 1 sits at their own seat = 4 * 3! (choose 1 out of 4 teachers, force them to sit at their own desk, and then let the others sit anywhere)
# of ways at least 2 sit at their own seat = 6 * 2! (choose 2 out of 4 teachers, force them to sit at their own desk, and then let the others sit anywhere)
# of ways at least 3 sit in their own seat = 4 * 1! (choose 3 out of 4 teachers, force them to sit at their own desk, and let the last one sit anywhere. Of course, that happens to be their own seat anyway, but the inclusion/exclusion principle is just designed to work like this.)
# of ways all 4 sit in their own seat = 1.

\[ \therefore \text{Ans: }4! - 4\times3! + 6\times2! - 4\times1! + 1 =9.\]

[PolySquared]

Help please

Let C and D be points with position vectors c and d respectively. If |c| = 5 and |d| = 7  and c dot d = 4, find the absolute value of vector CD