differentiation w trig (a level maths)

[gusty mantle]

uhm idk what to do after this

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[gusty mantle]

ignore my writing btw

[marble ember]

Don't let the clickbait title intimidate you

Note that $$a \sin x+b \cos x=R \sin (x+\alpha)$$ for some constants $R, \alpha$

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Civil Service Pigeon

[marble ember]

Since the maximum value of $y$ is $4$, we can let $R=4$

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Civil Service Pigeon

[marble ember]

So we have that $y=4 \sin (x+\alpha)$

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Civil Service Pigeon

[marble ember]

Since the curve passes through $\left(\frac{\pi}{3}, 2 \sqrt 3 \right)$, solving for $\alpha$ is trivial

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Civil Service Pigeon

[marble ember]

and then you can expand this using $$\sin(x+y)=\sin x \cos y+\sin y \cos x$$ to obtain the values of $a, b$

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Civil Service Pigeon

[marble ember]

Method 2: By Cauchy Schwarz, you can say that $$a \sin x +b \cos x \leq \sqrt{a^2+b^2}$$

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Civil Service Pigeon

[marble ember]

This gives you the system $$\begin{cases} \sqrt{a^2+b^2}=4 \ \frac{\sqrt 3}{2} a +\frac{1}{2} b=2 \sqrt 3 \end{cases}$$

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Civil Service Pigeon

[marble ember]

Method 3: Straight differentiation: $$y'=a \cos x -b \sin x =0$$ $$a - b \tan x=0$$

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Civil Service Pigeon

[marble ember]

$\tan x=\frac{a}{b} \implies \sin x=\frac{b}{\sqrt{a^2+b^2}}, \cos x=\frac{a}{\sqrt{a^2+b^2}}$

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Civil Service Pigeon

[marble ember]

It's basically another way to derive this

the finish is the same

If you want more approaches, lmk, but I think 3 is enough

[near abyss]

I think only the reverse is true. Otherwise the signs of sin(x) and cos(x) may also be opposite, depending on where x is, because a/b = (-a)/(-b).

[gusty mantle]

it rlly did LOL 😭

tysm tho

[gusty mantle]

this step is what i dont get, wtf is the sqrt(a² + b²)

[near abyss]

You can just use addition of harmonics, as usual.

[gusty mantle]

no but like what actually is it?

[near abyss]

Addition of harmonics allows you to express a sum of sines or cosines with the same frequency as just one function.
For example, a cos(ωt) + b sin(ωt) = c cos(ωt - φ), where c = √(a^2 + b^2), cos(φ) = a/c and sin(φ) = b/c. In other words, here φ = arg(a + bi).

[gusty mantle]

i meant the sqrt(a^2 + b^2)

[near abyss]

As I wrote above, it's the amplitude of the resulting function.

[gusty mantle]

oh okay ty

[keen hollow]

@gusty mantle has given 1 rep to @near abyss

[marble ember]

Yeah but b is clearly positive so

[near abyss]

What do you mean? Both a and b can be positive or negative in general.

[marble ember]

put x = 0 and you get y = b > 0

[near abyss]

I'm not talking about this exercise.

[marble ember]

ah I see

[wide pawn]

well for trig functions in the form of $asin(x)+bcos(x)$, $f(x) \in [-\sqrt{a^{2}+b^{2}},\sqrt{a^{2}+b^{2}}]$

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rev

[wide pawn]

and by given, u have √(a²+b²)=4

for next part, u have that the line passes thru the point (π/3, 2√3), so plug in these values into the asinx+bcosx=y which gives u 2 equations

solve for a, b

[near abyss]

I would probably do this exercise in a different way.
We can start with the form f(x) = A cos(ωx + φ). We obviously have A = 4 and ω = 1, so that becomes f(x) = 4cos(x + φ).
Next, we substitute the point.
4cos(π/3 + φ) = 2√(3)
cos(π/3 + φ) = √(3)/2
π/3 + φ = ±π/6 + 2πn, n ∈ ℤ
φ = -π/3 ± π/6 + 2πn, n ∈ ℤ
The period doesn't matter for phase.
φ = -π/3 ± π/6
f(x) = 4cos(x - π/3 ± π/6)
To determine which value to take, let's substitute x = 0.
f(0) = 4cos(-π/3 ± π/6) = 4(cos(-π/3)cos(π/6) ∓ sin(-π/3)sin(π/6)) = 4(cos(π/3)cos(π/6) ± sin(π/3)sin(π/6)) = √(3)(1 ± 1)
As we see from the graph that f(0) > 0, the value with the plus sign works. So:
f(x) = f(x) = 4cos(x - π/3 + π/6) = 4cos(x - π/6)
All that's left is to expand it. No addition of harmonics needed.

[gusty mantle]

this is a level maths

thats why it was easier for that before

and btw i am fourteen i just rlly like maths i dont do a level yet

[near abyss]

Sorry, don't understand what you mean.

[gusty mantle]

im from england

a level is what you do from 16 - 18

this is like the end of first year content

[near abyss]

Oh, wait.

[gusty mantle]

so i dont think ppl learn this method

[near abyss]

Do you mean "A-level"?

[gusty mantle]

i am fourteen learning a level maths for fun

[gusty mantle]

yeah 😭😭

[near abyss]

Ohh. I see.

[gusty mantle]

sorry lmfao icba to get the hyphen

[near abyss]

Really? Hm... Quite odd. I'd say this is the simplest approach, as this doesn't require you to know addition of harmonics.

How would you approach this, then?

[gusty mantle]

what civil service pigeon said

but i got stuck w addition of harmonics bc ive NEVER heard of that

[near abyss]

That is basically addition of harmonics, though. How would you know the resulting amplitude otherwise?

[gusty mantle]

yeah i didnt know, so i came here