Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

appleibeats · Sep 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I substituted to check that I am correct. But am unsure what you mean by consider whether I have missed any solutions. Is the answer NO.

Ambility · Sep 16, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
I substituted to check that I am correct. But am unsure what you mean by consider whether I have missed any solutions. Is the answer NO.

I don't think you've missed any solutions.

braintic · Sep 17, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
I substituted to check that I am correct. But am unsure what you mean by consider whether I have missed any solutions. Is the answer NO.

If sin2x = 1, then 2x = 90 OR 270 (0 < 2x < 360)

x = 45, 135 (0 < x < 180)

appleibeats · Sep 17, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

But at 270 is not sin2x = -1

So its not included??

appleibeats · Sep 17, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Express cos3theta in terms of powers of cos theta

rand_althor · Sep 17, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Express cos3theta in terms of powers of cos theta

cos(3x)
= cos(2x+x)
= cos(2x)cos(x) - sin(2x)sin(x)
= [2cos²(x)-1]cos(x) - [2sin(x)cos(x)]sin(x)
= 2cos³(x) - cos(x) - 2sin²(x)cos(x)
= 2cos³(x) - cos(x) - 2cos(x)[1 - cos²(x)]
= 2cos³(x) - cos(x) - 2cos(x) + 2cos³(x)
= 4cos³(x) - 3cos(x)

appleibeats · Sep 17, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Let sinA = 3/5 and cosB = 3/10^1/2

i)Evaluate sinB

ii) Hence, given that A and B are both acute, prove that A = 2B without evaluating A or B .

VBN2470 · Sep 17, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For part (a) use, $\sin{B}=\sqrt{1-\cos^{2}{B}}$ . Use a similar method for (b) as well.

braintic · Sep 17, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
But at 270 is not sin2x = -1

So its not included??

Yes - my mistake.

appleibeats · Sep 18, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

VBN2470 said:
For part (a) use, $\sin{B}=\sqrt{1-\cos^{2}{B}}$ . Use a similar method for (b) as well.

Can $\sin{B}=\sqrt{1-\cos^{2}{B}}$ be + or - as you take the square root.

appleibeats · Sep 18, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

SOlve the following equation for 0 < or equal to x < or equal to 360 using the t - formula

secx + sinx = cosx

Do I need to simplify before using the t formula? I didn't and my working is now getting complicated and I think I should have??

braintic · Sep 18, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
SOlve the following equation for 0 < or equal to x < or equal to 360 using the t - formula

secx + sinx = cosx

Do I need to simplify before using the t formula? I didn't and my working is now getting complicated and I think I should have??

Once you multiply to remove fractions then simplify, it is actually very uncomplicated.
The only trick is at the last step.

appleibeats · Sep 18, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

so i multiplied by cosx

to get

1 + sinxcosx = cos^2 x

1 + sinxcosx = 1 -sin^2 x

sin^2 x + sinxcosx = 0

The I used the t formula

and get

4t^2 - 2t^3 + 2t / ( 1 + t^2)^2 = 0

Not sure where to go from here??

Drongoski · Sep 18, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

How about:

sin² x + sinx cosx = 0

.: sin x (sin x + cos x) = 0

.: either sin x = 0 in which case x = . . .

or sin x + cos x = 0 ==> sin x = - cos x ==> sin x/cos x = tan x = -1

in which case x = . . .

appleibeats · Sep 19, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

so sinx = 0

x = 0 , 180

sinx + cosx = 0

x = 135

But the question asked to use the t formula . So is it still right??

Drongoski · Sep 19, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
so sinx = 0

x = 0 , 180, 360

sinx + cosx = 0

x = 135, 315

But the question asked to use the t formula . So is it still right??

If the question asks to use the t-formula (overlooked that), then you have to do it that way.

braintic · Sep 19, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
so i multiplied by cosx

to get

1 + sinxcosx = cos^2 x

1 + sinxcosx = 1 -sin^2 x

sin^2 x + sinxcosx = 0

The I used the t formula

and get

4t^2 - 2t^3 + 2t / ( 1 + t^2)^2 = 0

Not sure where to go from here??

I meant multiply after applying the t-results.

dan964 · Sep 19, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

note that sec x = 1/cos x, then put the t-results.
and then get rid of fractions and solve the quadratic.
and don't forgot that you are solving for t which tan (theta/2) meaning the range of values will change.
(latex is playing up)

appleibeats · Sep 19, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

so I let t = tan ( x/2)

Domain 0< x/2 < 180 ( less than and equal to)

then use t formula

1 + t^2 / 1 - t^2 + 2t / 1 + t^2 = 1 - t^2/ 1 + t^2

1 + t^2 + 2t / 1 - t^4 = 1 - t^2 / 1 + t^2

Then I cross multiply and expand and simplify to get

t^6 - 2t^4 - 2t^3 - 3t^2 - 2t = 0

Not sure where to go from here.

Looks terribly wrong.

rand_althor · Sep 19, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$\begin{align*}\sec{x}+\sin{x}&=\cos{x} \\\frac{1+t^2}{1-t^2}+\frac{2t}{1+t^2}&=\frac{1-t^2}{1+t^2} \\\textup{Multiply through by }(1-t^2)&(1+t^2) \\(1+t^2)^2+2t(1-t^2)&=(1-t^2)^2 \\(1+t^2)^2-(1-t^2)^2+2t(1-t^2)&=0 \\\textup{Since }a^2-b^2&=(a-b)(a+b) \\(1+t^2+1-t^2)(1+t^2-(1-t^2))+2t(1-t^2)&=0 \\2(2t^2)+2t(1-t^2)&=0 \\2t(2t+1-t^2)&=0 \\t(t^2-2t-1)&=0 \\\end{align*}$

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