PAT Solutions.

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Reply 20

Fing4

Original post by Maths_Lover

2010 - Q1(i)

Spoiler

\sin

as it is the easiest given the coefficients we have.

\sin 3x - \sqrt{3} \cos 3x \equiv R\sin (3x - \alpha )

Use the addition formulae to multiply out the RHS:

\sin 3x - \sqrt{3} \cos 3x \equiv R(\sin 3x\cos \alpha - \cos 3x\sin \alpha )

\sin 3x - \sqrt{3} \cos 3x \equiv R\sin 3x\cos \alpha - R\cos 3x\sin \alpha

Comparing coefficients, it is clear that:

R\cos \alpha = 1\ \ \ \ \ \ \ \ \ R\sin \alpha = \sqrt{3}

\Rightarrow \tan \alpha = \sqrt{3}

From special triangles,

\alpha = \dfrac{\pi }{3}

.

Add the two equations together to obtain a third:

R\cos \alpha + R\sin \alpha = 1 + \sqrt{3}

Square both sides and factorise the LHS:

R^2(\cos ^2\alpha + \sin ^2\alpha ) = (1 + \sqrt{3} )^2

However

\cos ^2\alpha + \sin ^2\alpha \equiv 1

\therefore R^2 = (1 + \sqrt{3} )^2

R = 1 + \sqrt{3}

So,

\sin 3x - \sqrt{3} \cos 3x \equiv (1 + \sqrt{3} )\sin (3x - \dfrac{\pi }{3} ) = 0

(1 + \sqrt{3} )\sin (3x - \dfrac{\pi }{3} ) = 0

\sin (3x - \dfrac{\pi }{3} ) = 0

(3x - \dfrac{\pi }{3} ) = \arcsin 0

Looking at the specified range to obtain solutions within:

0 \leq x \leq \pi

However, we have 3x, so multiplying by 3 gives

0 \leq 3x \leq 3\pi

.

Now we can write down the solutions of

(3x - \dfrac{\pi }{3} ) = \arcsin 0

that would be within the new range when we add

\dfrac{\pi }{3}

to both sides:

(3x - \dfrac{\pi }{3} ) = 0, \pi , 2\pi \ \ \

(don't forget that there is more than one value for

\arcsin 0

in the new range!)

Adding

\dfrac{\pi }{3}

to both sides then dividing both sides by 3, we obtain:

x = \dfrac{\pi }{9} ,\dfrac{4\pi }{9} ,\dfrac{7\pi }{9}

R is 2 here.

R^2=1^2 + sqrt(3)^2 =1+3=4
R=sqrt4=2

Edit: you don't add the equations then square them. You square the equations THEN add them

(edited 12 years ago)

Reply 21

Maths_Lover

Original post by Fing4

R is 2 here.

R^2=1^2 + sqrt(3)^2 =1+3=4
R=sqrt4=2

Just saw your edit. Hang on a min.

(edited 12 years ago)

Reply 22

Fing4

Original post by Maths_Lover

I see what you are getting at, however we are squaring the whole of both sides:

1^2 + (\sqrt{3} )^2 \not= (1+\sqrt{3} )^2

So R is not 2...

Forgive my rudeness but you are wrong.

Rsinx=1
Rcosx=sqrt3

R^2sin^2x=1
R^2cos^2x=3

R^2(cos^2x+sin^2x)=4
R=2

Edit: If you still don't believe me, try wolfram alpha http://www.wolframalpha.com/input/?i=sin3x-sqrt3cos3x, amplitude of the graph is 2, hence R=2

(edited 12 years ago)

Reply 23

Maths_Lover

Original post by Fing4

R is 2 here.

R^2=1^2 + sqrt(3)^2 =1+3=4
R=sqrt4=2

Edit: you don't add the equations then square them. You square the equations THEN add them

I see what you mean now. My bad. :lol:

I shall edit my solution accordingly. Thanks for that. :smile:

Reply 24

Fing4

Original post by Maths_Lover

I see what you mean now. My bad. :lol:

I shall edit my solution accordingly. Thanks for that. :smile:

Haha don't worry, you're welcome

Reply 25

Maths_Lover

Original post by Fing4

Haha don't worry, you're welcome

Better to find these things out before the exam. :colone:

I hadn't realised you do it the other way round to begin with and got confused as to what you were on about before I saw your edit.

Reply 26

matthew769

2006 Q10 -

Spoiler

Reply 27

bananarama2

2010-Q4

Spoiler

Reply 28

bananarama2

2010-Q5

Spoiler

(edited 12 years ago)

Reply 29

bananarama2

2010-Q25

i)

Spoiler

ii)

Spoiler

iii)

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iv)

Spoiler

vi)

Spoiler

vii)

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viii)

Spoiler

(edited 12 years ago)

Reply 30

Maths_Lover

2010 - Q7

Spoiler

Reply 31

matthew769

2008 Q1

Spoiler

Reply 32

matthew769

2008 Q2

Spoiler

(edited 12 years ago)

Reply 33

Maths_Lover

2010 - Q8

Spoiler

(edited 12 years ago)

Reply 34

Maths_Lover

Original post by matthew769

2008 Q2

Spoiler

\sqrt{whatever you want square rooted}

in LaTex, leaving a space after the curly brackets. :cute:

Reply 35

bananarama2

Original post by matthew769

2008 Q4

Spoiler

There are several things wrong with that....

What if x=5, the denominator would be a negative. When you times by a negative in inequalities, The sign flips....