Help with Trig induction (1 Viewer)

MAK03 · Mar 11, 2022

I need help with basically every question on this sheet, if there are any questions you know how to answer please send

=)(= · Mar 11, 2022

i can help with q2 ill send the solution through in like 5 mins

ExtremelyBoredUser · Mar 11, 2022

=)(= · Mar 11, 2022

found a solution yet i doubt its validity

they have not proven the n=k+1 case, they have just proven the assumption

ExtremelyBoredUser · Mar 11, 2022

Q3

I'm sure there's an easier way (for sure) but this was the most apparent from the start (and its 11 pm I don't want to think rn haha). The main idea is that I tan'd both sides of the claims and I would use the tan(A+B) or tan(A-B) formula to produce a simpler expression. The last part is just algebra and its just trying to find a way to make it like the RHS expression so its just factoring.

5uckerberg · Mar 12, 2022

MAK03 said:
I need help with basically every question on this sheet, if there are any questions you know how to answer please send

Q5

You start off with
$\frac{\cos{y}-\cos\left(y+2x\right)}{2\sin{x}}=\sin\left(y+x\right)$

What you have to do here is that since there are no obvious inroads into the Q you have to expand the $\cos\left(y+2x\right)$ term.

Thus, it becomes

$\frac{\cos{y}-\left(\cos{y}\cos{2x}-\sin{y}\sin{2x}\right)}{2\sin{x}}$
At this point, you need to have a sharp eye and notice the terms $\cos{2x} \land \sin{2x}$ . They will be your angels.

To show it recall your trig identities

$\frac{\cos{y}-\cos{y}\left(\cos^{2}x-\sin^{2}x\right)+\sin{y}\left(2\sin{x}\cos{x}\right)}{2\sin{x}}$ .
Using the fact that $\sin{2x}=2\sin{x}\cos{x}$ and $\cos{2x}=\cos^{2}x-\sin^{2}x$

Looks like a mess right?

Well, tidy it up

If you have finished your chores then you should have

$\frac{\cos{y}\left(1-\cos^{2}x-\sin^{2}x\right)+2\sin{y}\sin{x}\cos{x}}{2\sin{x}}$

$\frac{2\sin^{2}x\cos{y}+2\sin{y}\cos{x}\sin{x}}{2\sin{x}}$

Removing $2\sin{x}$ we will obtain

$\sin{x}\cos{y}+\cos{x}\sin{y}=RHS$

QED

b) Start off with n=1 because the condition is for all positive integral values. Note integral simply means integer.

$\sin{x}+\sin{3x}+\sin{5x}+...+\sin\left(2n-1\right)x=\frac{1-\cos\left(2nx\right)}{2\sin{x}}$

Applying what we said at the start we will have

$\sin{x}=\frac{1-\cos{2x}}{2\sin{x}}$
Well, competance in your trig identities is key.
$\cos{2x}=1-2\sin^{2}x$
$\sin{x}=\frac{1-1+2\sin^{2}x}{2\sin{x}}\rightarrow{\sin{x}=\frac{2\sin^{2}x}{2\sin{x}}$

QED

For the n=k case

$\sin{x}+\sin{3x}+\sin{5x}+...+\sin\left(2k-1\right)x=\frac{1-\cos\left(2kx\right)}{2\sin{x}}$ $\left(\mathbf{Assumption}\right)$

RTP for n=k+1 case

$\sin{x}+\sin{3x}+\sin{5x}+...+\sin\left(2k-1\right)x+\sin\left(2k+1\right)x=\frac{1-\cos\left(\left(2\left(k+1\right)\right)x\right)}{2\sin{x}}$
$\sin{x}+\sin{3x}+\sin{5x}+...+\sin\left(2k-1\right)x+\sin\left(2k+1\right)x=\frac{1-\cos\left(2kx+2x\right)}{2\sin{x}}$

Using $\left(\mathbf{Assumption}\right)$

The n=k+1 case becomes

$\frac{1-\cos\left(2kx\right)}{2\sin{x}}+\sin\left(2k+1\right)x=\frac{1-\cos\left(2kx+2x\right)}{2\sin{x}}$

Put the LHS under the same denominator

$\frac{1-\cos\left(2kx\right)+2\sin{x}\sin\left(2k+1\right)x}{2\sin{x}}$

Is this just

$\frac{1-\cos\left(2kx\right)+\cos\left(2kx\right)-\cos\left(2k+2\right)x}{2\sin{x}}$

Using the fact that $2\sin{x}\sin\left(y+x\right)=\cos{y}-\cos\left(y+2x\right)$

After finishing it we will have

$\frac{1-\cos\left(2k+2\right)x}{2\sin{x}}=RHS$
$\frac{1-\cos\left(2\left(k+1\right)\right)x}{2\sin{x}}=RHS$

QED. Then you can conclude your statement.

5uckerberg · Mar 12, 2022

Is there a function from LaTeX where I can cross out sections of the working through simplification?

5uckerberg · Mar 12, 2022

Q6i

$\sin\left(A+B\right)-\sin\left(A-B\right)=\sin{A}\cos{B}+\cos{A}\sin{B}-\left(\sin{A}\cos{B}-\cos{A}\sin{B}\right)$
That is just
$2\cos{A}\sin{B}$

ii

Q for reference

$\cos{x} + \cos{2x} + \cos{3x} + ... + \cos{nx} = \frac{\sin\left(n+\frac{1}{2}\right)x-\sin\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)}$

For all positive integers n

Well, let's start with n=1

$\cos{x}=\frac{\sin\left(\frac{3}{2}\right)x-\sin\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)}$
Using part i is this just

$\frac{2\cos{x}\sin\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)}=\cos{x}$ .
Thus, it is true for n=1.

For n=k

$\cos{x} + \cos{2x} + \cos{3x} + ... + \cos{kx} = \frac{\sin\left(k+\frac{1}{2}\right)x-\sin\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)} \left(\mathbf{Assumption}\right)$

RTP for n=k+1 case

$\cos{x} + \cos{2x} + \cos{3x} + ... + \cos{\left(k+1\right)x} = \frac{\sin\left(k+1+\frac{1}{2}\right)x-\sin\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)}$

Using $\left(\mathbf{Assumption}\right)$

$\frac{\sin\left(k+\frac{1}{2}\right)x-\sin\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)} + \cos{\left(k+1\right)x} = \frac{\sin\left(k+1+\frac{1}{2}\right)x-\sin\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)}$

with the LHS write with a common denominator

$\frac{\sin\left(k+\frac{1}{2}\right)x-\sin\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)} + \frac{2\sin\left(\frac{x}{2}\right)\cos{\left(k+1\right)x}}{2\sin\left(\frac{x}{2}\right)} = \frac{\sin\left(k+1+\frac{1}{2}\right)x-\sin\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)}$

Okay, now what?

Use part i which you have proven already

$\sin\left(A+B\right)-\sin\left(A-B\right)=2\cos{A}\sin{B}$

Notice that $A=\left(k+1\right)x, B=\frac{x}{2}$

$\frac{\sin\left(k+\frac{1}{2}\right)x-\sin\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)} + \frac{2\sin\left(\frac{x}{2}\right)\cos{\left(k+1\right)x}}{2\sin\left(\frac{x}{2}\right)}$

becomes

$\frac{\sin\left(k+\frac{1}{2}\right)x-\sin\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)} + \frac{\sin\left(k+1+\frac{1}{2}\right)x-\sin\left(k+\frac{1}{2}\right)x}{2\sin\left(\frac{x}{2}\right)}$

Cleaning it up we will have

$\frac{\sin\left(k+1+\frac{1}{2}\right)x-\sin\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)}=RHS.$

There you can finish it off.

5uckerberg · Mar 12, 2022

Q4

Q for reference

$\tan{x}+2\tan{2x}+4\tan{4x}+...+2^{n-1}\tan2^{n-1}x=\cot{x}-2^{n}\cot\left(2^{n}x\right)$ for $n\geq{1}$

Start it off let n=1

$\tan{x}=\cot{x}-2\cot{2x}$
$\tan{x}=\frac{1}{\tan{x}}-\frac{2}{\tan{2x}}$
$\tan{x}=\frac{\tan{2x}-2\tan{x}}{\tan{x}\tan{2x}}$
$\tan{x}=\frac{\frac{2\tan{x}}{1-\tan^{2}x}-2\tan{x}}{\tan{x}\frac{2\tan{x}}{1-\tan^{2}x}}$
$\frac{2\tan{x}-2\tan{x}+2\tan^{3}x}{2\tan^{2}x}=\tan{x}$

Now we prove it is true for n=k

$\tan{x}+2\tan{2x}+4\tan{4x}+...+2^{k-1}\tan2^{k-1}x=\cot{x}-2^{k}\cot\left(2^{k}x\right)$ $\left(\mathbf{Assumption}\right)$

RTP for n=k+1

$\tan{x}+2\tan{2x}+4\tan{4x}+...+2^{k}\tan2^{k}x=\cot{x}-2^{k+1}\cot\left(2^{k+1}x\right)$
Inserting the $\left(\mathbf{Assumption}\right)$ on the LHS we will have
$\cot{x}-2^{k}\cot\left(2^{k}x\right)+2^{k}\tan2^{k}x=\cot{x}-2^{k+1}\cot\left(2^{k+1}x\right)$
See you later $\cot{x}$
$-2^{k}\cot\left(2^{k}x\right)+2^{k}\tan2^{k}x=-2^{k+1}\cot\left(2^{k+1}x\right)$
$2^{k}\tan2^{k}x=2^{k}\cot\left(2^{k}x\right)-2^{k+1}\cot\left(2^{k+1}x\right)$
$2^{k}\tan2^{k}x=\frac{2^{k}}{\tan\left(2^{k}x\right)}-\frac{2^{k+1}}{\tan\left(2^{k+1}x\right)}$
$2^{k}\tan2^{k}x=\frac{2^{k}\tan\left(2^{k+1}x\right)-2^{k+1}\tan\left(2^{k}x\right)}{\tan\left(2^{k}x\right)\tan\left(2^{k+1}x\right)}$

Focus on the RHS
$\frac{2^{k}\tan\left(2^{k+1}x\right)-2^{k+1}\tan\left(2^{k}x\right)}{\tan\left(2^{k}x\right)\tan\left(2^{k+1}x\right)}=\frac{2^{k}\left(\tan\left(2^{k+1}x\right)-2\tan\left(2^{k}x\right)\right)}{\tan\left(2^{k}x\right)\tan\left(2^{k+1}x\right)}$

Recall that $\tan\left(2^{k+1}x\right)=\tan\left(2^{k}x+2^{k}x\right)$

In that case, we will have

$\frac{2^{k}\left(\frac{2\tan\left(2^{k}x\right)}{1-\tan^{2}\left(2^{k}x\right)}-2\tan\left(2^{k}x\right)\right)}{\tan\left(2^{k}x\right)\frac{2\tan\left(2^{k}x\right)}{1-\tan^{2}\left(2^{k}x\right)}$

Next,

$\frac{2^{k}\left(2\tan\left(2^{k}x\right)-2\tan\left(2^{k}x\right)+2\tan^{3}\left(2^{k}x\right)\right)}{2\tan^{2}\left(2^{k}x\right)}$
$\frac{2^{k}\left(2\tan^{3}\left(2^{k}x\right)\right)}{2\tan^{2}\left(2^{k}x\right)}$
$2^{k}\tan\left(2^{k}x\right)$

Adding $\cot{x}$ we have just proven the n=k+1 statement.

The conclusion is up to you.

5uckerberg · Mar 12, 2022

Is there someone who can finish Q6iii. Seems like a lot of mental sweat is required.

5uckerberg · Mar 12, 2022

=)(= said:
found a solution yet i doubt its validity
View attachment 35155
they have not proven the n=k+1 case, they have just proven the assumption

I reckon that is what should be done because they used the identity from part i.

5uckerberg · Mar 12, 2022

Q1)
Show that $\sin\left(f\left(x\right)\right)+\cos\left(f\left(x\right)\right)=\sqrt{2}\sin\left(f\left(x\right)+\frac{\pi}{4}\right)$

Q for reference

If you had learnt your Algebra from Year 7 there is an intrinsic technique here. Remember x by itself is simply 1 times x

$\sin\left(f\left(x\right)\right)+\cos\left(f\left(x\right)\right)=1\times{\sin\left(f\left(x\right)\right)}+1\times{\cos\left(f\left(x\right)\right)}$
Here, we will be using the fact that this is an auxiliary angle. I like to call it basically wrapping a present because you are going from $\sin{x}\cos{y}+\cos{x}\sin{y}=\sin\left(x+y\right)$ .
Well

$R\cos{\alpha}=1$
$R\sin{\alpha}=1$
$\tan{\alpha}=1, \alpha=\frac{\pi}{4}$
$R^{2}\left(\cos^{2}x+\sin^{2}x\right)=2$
$R^{2}=2$
$R=\sqrt{2}$ because the negative version requests that you have -1 and $-\frac{\pi}{4}$
Therefore,
$R\left(\sin{f(x)}\cos{\alpha}+\cos{f(x)}\sin{\alpha}\right)=R\sin\left(f(x)+\alpha\right)$
Sub in all the values we have found and you have finished part a

bi) Given that $y=e^{x}\sin{x}$
Differentiate using product rule and we will have
$\frac{dy}{dx}=e^{x}\left(\sin{x}+\cos{x}\right)$ .
What do we do next, recall part a
$\sin\left(f\left(x\right)\right)+\cos\left(f\left(x\right)\right)=\sqrt{2}\sin\left(f\left(x\right)+\frac{\pi}{4}\right)$
we have an easier version from $\frac{dy}{dx}$ .
Therefore, $\frac{dy}{dx}=\sqrt{2}e^{x}\sin\left(x+\frac{\pi}{4}\right)$

5uckerberg · Mar 12, 2022

ExtremelyBoredUser said:
Q3

I'm sure there's an easier way (for sure) but this was the most apparent from the start (and its 11 pm I don't want to think rn haha). The main idea is that I tan'd both sides of the claims and I would use the tan(A+B) or tan(A-B) formula to produce a simpler expression. The last part is just algebra and its just trying to find a way to make it like the RHS expression so its just factoring.

@ExtremelyBoredUser
Given your post that is why I named these files like this.

5uckerberg · Mar 13, 2022

Q6iii

Yesterday in my sleep I had a very good idea to finish the question.

On the LHS use the fact that $\cos\left(A-B\right)+\cos\left(A+B\right)=2\cos{A}\cos{B}$ and then slowly close up the present as you repeat that process three times.

ExtremelyBoredUser · Mar 13, 2022

5uckerberg said:
@ExtremelyBoredUser
Given your post that is why I named these files like this.

I legit slept halfway through and then I realised I didn't even finish working so I rushed it.. but yes ExtremelyTiredUser is more fitting nowadays. If only I could change usernames.

stupid_girl · Mar 14, 2022

5uckerberg said:
Q6iii

Yesterday in my sleep I had a very good idea to finish the question.

On the LHS use the fact that $\cos\left(A-B\right)+\cos\left(A+B\right)=2\cos{A}\cos{B}$ and then slowly close up the present as you repeat that process three times.

There is the word "hence" so you may consider using (ii).

Aftet applying (ii), you get a difference of sine in the numerator and you can now apply (i).
sin(9x+8x)-sin(9x-8x)

If you are familiar with double angle formula, you may recognize sin8x=8 sinx cosx cos2x cos4x.

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Help with Trig induction (1 Viewer)

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