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secondary 4 | A Maths
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glittr
Glittr

secondary 4 chevron_right A Maths chevron_right Singapore

hi pls help with qn in the picture.
i tried but i get “undefined” answer .

Date Posted: 4 years ago
Views: 650
Eric Nicholas K
Eric Nicholas K
4 years ago
y = [cos (3x)]^2

In your online calculator, you must put them as such. Do not put the function y = cos (3x)^2, because the online calculator recognises this as cos (9x^2) which is entirely different. However, y = (cos 3x)^2 is perfectly ok.

y = [cos (3x)]^2
dy/dx
= 2 [cos (3x)]^1 * d/dx [cos (3x)] by the chain rule
= 2 cos (3x) * [- sin (3x)] * d/dx (3x) by a second chain rule
= -2 cos (3x) sin (3x) * 3
= -6 sin (3x) cos (3x)

When x = pi/2 (which is 90 degrees)

dy/dx
= -6 sin (3pi/2) cod (3pi/2)
= -6 (-1) (0)
(because sin 270 = -1 and cos 270 = 0, this can be seen by drawing the most basic sine and cosine graphs)
= 0

denoting that the gradient of the curve at (pi/2, 0) is 0.
Eric Nicholas K
Eric Nicholas K
4 years ago
Without using differentiation, I can intuitively tell that the gradient at that point must be zero.

You can think of y = [cos (3x)]^2 as a perfect square y = Z^2 where Z = cos (3x). As we all know, the minimum possible value of a perfect square is 0, so the curve cannot possibly have a negative y-value.

Hence, the graph must turn at the lowest value of 0, and at this turning point the gradient is zero.

The graph has a maximum value of 1, because the original un-squared function cos (3x) has a defined interval of -1 <= value <= 1 and squaring this interval gets us 0 <= value <= -1

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Tan Hui Juan
Tan Hui Juan's answer
2 answers (Tutor Details)
1st
Check your differentiation or whether u change to rad mode in your calculator.
Eric Nicholas K
Eric Nicholas K
4 years ago
It’s -6 cos 3x sin 3x. You missed out a negative sign.

Having said that, the missing negative sign should not affect the solution.
glittr
Glittr
4 years ago
hi i checked the mode and used online diff calculator to confirm . but still get calculator error . can help pls
glittr
Glittr
4 years ago
hi all , thanks for input . i just realised i uploaded the wrong qn . pls help see latest pic :)
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glittr
Glittr's answer
59 answers (Tutor Details)
updated qn
Eric Nicholas K
Eric Nicholas K
4 years ago
It’s not that straightforward.

Strictly speaking, it’s not just the sec^2 3x which is invalid when x = pi/6. The tan^2 3x in the denominator is also technically undefined when x = pi/6.

The fact that the point pi/6 does exist in the original graph means we need to somehow modify our expression for dy/dx (by the way, your denominator should be the whole thing squared).

I will have a look at this again when I am more free.
Eric Nicholas K
Eric Nicholas K
4 years ago
I just observed. Your expression for dy/dx can further be simplified. We return our denominator to (sec^2 3x)^2, and then simplify to -6 Tan 3x divided by sec^2 3x, which is -6 tan 3x cos^2 3x.

This is where you can set dy/dx to 0.

Rather, this second approach is actually quite invalid because your introduction of the secant renders the pi/6 solution inadmissible (since secant of 3 pi/6 is undefined).
Eric Nicholas K
Eric Nicholas K
4 years ago
In short, this approach fails because of the unnecessary introduction of the secant function which causes pi/6 to be invalid.
glittr
Glittr
4 years ago
is using secant considered an “unlucky “ method ? like we wouldn’t know in the beginning if it could work or fail
Eric Nicholas K
Eric Nicholas K
4 years ago
It’s not just secant which falls prey to this approach. Any approach which causes a denominator to be formed usually creates this issue.

Take for example the simplest case.

Solve x - 3 = 0.

One can easily say that x = 3.

But why not we do this?

x - 3 = (x - 3)^2 / (x - 3)

They are the same, but the unnecessary introduction of the denominator causes the same solution x = 3 to be rejected, in view that denominators cannot be equated to zero.

For the same idea, secant fails miserably here.
Eric Nicholas K
Eric Nicholas K
4 years ago
Logarithmic functions also have this problem. This is the reason why the solutions to

ln x^2 = 0

and

2 ln x = 0

are different, even though both equations are immediately interchangeable by the logarithmic rules. The solutions are x = -1 and x = 1, but all because ln (-1) is undefined, we reject the solution x = -1 for the 2 ln x = 0 case.

So if a question is posed as ln x^2 = 0, taking the approach 2 ln x = 0 instead of converting to the index form x^2 = 1 causes us to lose the negative solution which would have otherwise been valid.
Eric Nicholas K
Eric Nicholas K
4 years ago
One more.

Solve 2x = 0.

Method 1: Correct

Dividing both sides by 2, x = 0.

Method 2: Flawed

2 / (1/x) = 0
1/x = 2/0
= infinity
x = 1/infinity
= 0

The secant idea closely follows this example.
glittr
Glittr
4 years ago
wow thanks for the detailed explanation . i never knew that introducing denominators are considered dangerous