Year 13 Maths Help Thread

$\sin 2\theta = 2\sin \theta \cos \theta$. You know what $\cos \theta$ is so now you need only try and find $\sin \theta$, there's several things you can do here. One is draw a standard basic 3-4-5 triangle, another is use the Pythagorean identity. See if you make any progress.

Draw a right angled triangle. Mark one of the angles as $\theta$. Then you know the adjacent side is length 3 and hypotenuse is 5. You can use Pythagoras to work out the third side, thus you know $\sin(\theta)$

Thanks, yeah i got it now

What about cos( 2 theta)?

I found theta then subbed into 2cos^2 (theta) - 1 to get -7/25

but the mark scheme says 9/25 - 16/25

Thanks, yeah i got it now

What about cos( 2 theta)?

I found theta then subbed into 2cos^2 (theta) - 1 to get -7/25

but the mark scheme says 9/25 - 16/25

You don't even need to find $\theta$ itself since you know the values of $\sin(\theta)$ and $\cos(\theta)$

Okay, 9/25 - 16/25 = -7/25 so you're not wrong. They've used the identity in terms of sine and cosine for $\cos(2\theta)$ whereas you used the identity entirely in terms of $\cos(\theta)$.

Yeah i see it now because you already know sin theta and cos theta.

Please could you help me with 9bii, i tried to expand then differentiate but i didnt get the right answer

Correct approach. Show your working?

Sure, its a little messy but i understand it


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I meant Q9


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Oops i crossed it out whilst i was trying to do it. But this is what i did:

That's correct. You can express $e^{-\frac{t}{4}}$ in terms of $h$ if you refer to the model's equation.

Ah yeah, thanks

So would this get full marks?

Ah yeah, thanks

So would this get full marks?

Yep

Please could you explain 2c)ii) i dont really understand what im supposed to do.

For part a i know that the ranges would be IxI < 1 and IxI < 1/3

But im unsure of how you work it out when 2 binomial expansions are added/subtracted etc.

Please could you explain 2c)ii) i dont really understand what im supposed to do.

For part a i know that the ranges would be IxI < 1 and IxI < 1/3

But im unsure of how you work it out when 2 binomial expansions are added/subtracted etc.

Do you understand how we have the two inequalities of $\vert x \vert < 1$ and $\vert 3x \vert < 1$ from $(1+x)^{-1}$ and $(1+3x)^{-1}$ respectively??

So an expression where both are involved will only be valid for $x$ which satisfies BOTH inequalities.

In this diagram I label the region where $\vert x \vert < 1$ in red and $\vert 3x \vert < 1$ in green. For $x$ to satisfy both, you're looking where these two regions overlap. Obviously, this will be $\vert 3x \vert < 1$. Another way to think about it is that all values satisfying $\vert 3x \vert < 1$ are contained within $\vert x \vert < 1$ so the former is a stricter region, thus we choose it.

I dont really understand why but ive memorised it as the mod of x (from the expansion) is less than 1. Then rearrange so I3xI < 1 becomes IxI < 1/3

Yeah i see why IxI < 1/3 - its because it satisfies both IxI < 1 and IxI < 1/3 right?

Yep

Yep

Also i did one of those 'show that dy/dx' questions but im still not sure if i would get the marks. Its question 4c)i

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