STEP Tricks and Hacks Thread

Use the fact that some trigonometric functions are shifts of another, common examples include $\sin \theta = \cos (\pi/2 - \theta)$, $\cot \theta = \tan (\pi/2 - \theta)$ and the likes. This comes in useful and saves a lot of time.

Say you were given $\csc \left(\theta + \frac{\pi}{3}\right) = \sec \left(\theta - \frac{\pi}{6}\right)$ - this could be an 8/9/10mark A-Level question easily; want a three liner way of doing it?

The equation is equivalent to $\displaystyle \sin \left(\theta + \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{2} - \theta + \frac{\pi}{6}\right) = \sin \left(\frac{2\pi}{3} - \theta\right)$

which now becomes elementary to solve.

Extension to this, using a STEP question, show that $\sin A = \cos B \iff A = (4n+1)\frac{\pi}{2} \pm B$ - should be a two or three liner using this method.

This trick has rendered one or two STEP I questions completely inane, especially the $\cot \theta = \tan (\pi/2 - \theta)$ that's let me do a STEP I question in 4 minutes.

For the necessary condition, why is it that taking sine of the equation gives you -sin=cos whereas taking the cosine proves it for +B. I can see the symmetry but how do I know which one to use? Pic related will explain my question better tbh.

For the necessary condition, why is it that taking sine of the equation gives you -sin=cos whereas taking the cosine proves it for +B. I can see the symmetry but how do I know which one to use? Pic related will explain my question better tbh.




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Actually - I've spotted it right away, we have $\sin \left(A - (4n+1)\frac{\pi}{2}\right) = \sin \pm B$.

But this gives us $0 - \cos A = \sin \pm B$ which is the same thing as $-\cos A = \pm \sin B$ which certainly gives us $\cos A = \sin B$.

The $+B$ comes from the $\cos$ and then $-B$ comes from the $\sin$.

So remember when we have $-\cos A = \pm \sin B$ we have cos A is the same as (+sin B or -sin B) only one of those have to be true, not both.

Oh! Are you saying that I couldn't prove it because I was messing up my boolean logic/truth tables? So if we have +sin B or -sin B, it would be enough to say that cos A = sin B by just emitting the -sin B ?

Precisely.

You know how you feel really happy when you solve a math problem and learn something new? That is what I am feeling right now. Gonna start STEP III 2006, do one question and start M5 - I've been doing so much STEP today that I have forgotten I am really behind on Further Mathematics A-level.

Thank you again. Have fun, and I hope you enjoy your movie!

Say you wanted to differentiate a nasty function that might require multiple uses of the product rule or such, take logs!

Extension to this, using the same idea, roughly - attempt $\displaystyle \int_0^{\pi/2} \log \sin \theta \, \mathrm{d}\theta$

Why take logs when you can take natural logs? Or is this what you meant?



Similarly, you meant to say natural log ($\ln$) right?

At any level higher than A-level $\log$ and $\ln$ are interchangeable, in mathematics at least.

Why take logs when you can take natural logs? Or is this what you meant?



Similarly, you meant to say natural log ($\ln$) right?

Using implicit differentiation can make life easier.

Say you wanted to differentiate a nasty function that might require multiple uses of the product rule or such, take logs!

Let's say you have $f(x) = x^p(x+1)^q(x+2)^r$, do you really want to differentiate this normally? I'd hope you said no. What you'd want to do is:



and leaving your answer in this form, i.e: as $f(x)(\cdots)$ is quite useful in many a STEP I question that I've come across.

As an aside, here's another place where it's useful: $\frac{\mathrm{d}}{\mathrm{d}x} |f(x)|$, now the standard thing to do would be to consider intervals of $x$ that makes $f$ positive or negative and differentiate piecewise.

What'd I'd do is $y = |f(x)| \Rightarrow y^2 = f(x)^2 \Rightarrow 2yy' = 2f'(x)f(x) \Rightarrow y' = \frac{f'(x)f(x)}{|f(x)|}$ - this also lets you know where the function is not differentiable.

I'll edit it a nice problem for you lot to try in a bit.

Do you have one right now?

...

Although it's brief you might like to prove the product and quotient rules for suitably defined functions using this sort of technique.

As above - at any level above A-Level we have that $\log$ and $\ln$ are interchangeable.

Except in engineering, where they often use $\log = \log_{10}$ ...

And computer science, where sometimes $\log = \log_{2}$

And computer science, where sometimes $\log = \log_{2}$