Question

Mini guide. Tricks for figuring out some trig identities.

Answer 1

So let's just look at the unit circle to see if we can figure out why this identity: \[\Large \sin x \cos x = \frac{1}{2}\sin(2x)\] should be true. If we start going around we see that both are positive, so multiplying them together will give us a positive. Keep going around and notice in the second quadrant we have a negative sign on one of the functions. Go further still to the 3rd quadrant and we have negative. So now we see we're going from + to - twice in one rotation. That's our first good news, the frequency is twice as normal, so we have (2x) inside the trig function.

What next? Well we start out at 0 and we know we end up at 0 after pi/2. This is our hint that it's a sine function. So now what's the amplitude? We can be fancy and take the derivative and solve for when sinxcosx is at a maximum or we can just see that if it's 0 at x=0 and x=pi/2 then at x=pi/4 must be where the maximum of our sine function occurs. So that's just (sqrt2/2)^2 which is indeed 1/2. So there you have it. We can do similar things for cos^2x=1/2+1/2cos(2x) if we wish.

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Next trick I like is using euler's formula. Next post coming up!

Answer 2

\[\LARGE e^{i \theta}=\cos \theta + i \sin \theta\] This magic formula allows us to turn nearly any trig identity we want to find into a simple problem of exponents!

A quicky example is:
\[\LARGE e^{i (\theta+\phi)}=\cos( \theta + \phi)+ i \sin (\theta+ \phi)\]
can also be expressed as: \[\LARGE e^{i \theta}e^{i \phi}=[\cos( \theta )+ i \sin (\theta)][\cos( \phi )+ i \sin (\phi)]\] Now we equate the two and get: \[\Large \cos( \theta + \phi)+ i \sin (\theta+ \phi)\ =[\cos( \theta )+ i \sin (\theta)][\cos( \phi )+ i \sin (\phi)]\]

Distribute the right part out and look at the real and imaginary parts separately you get from this and you will see we have two familiar friends. Of course, we can do this more generally to get some really weird trig identities. =D

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next one up is fairly tame, but I think it should be said. It just involves the pythagorean identity.

Answer 3

\[\Large \sin^2(x)+\cos^2(x)=1\] Divide the whole thing through by sine or cosine to get: \[\Large \tan^2(x)+1=\sec^2(x)\]

Suppose you only remembered this fact but you want the sine squared one: \[\Large \cos^2(x)=\frac{1}{2}+\frac{1}{2}\cos(2x)\] Then plug this sucker in:\[\Large \sin^2(x)+[\frac{1}{2}+\frac{1}{2}\cos(2x)]=1\] And there you go, no fuss no muss.

That's all I really have, I don't know why I felt like posting this but I see these all the time and thought what the heck.

Answer 4

Thanks for sharing @Kainui :)

Answer 5

Thanks... @Kainui

Answer 6

That's hot, thanks kai!

Answer 7

My favorite that I discovered earlier this summer:

Suppose you want to integrate something like cos^7(x) what do you do? Well you can do this little shortcut: \[\large \cos^7( \theta)=\frac{1}{2^6}[1\cos(7 \theta)+7\cos(5 \theta)+21\cos(3 \theta)+35 \cos(\theta)]\]

Much easier to integrate, and the pattern is simple!

1) Divide by 2 raised to the same power -1.
2) Take the nth row of pascal's triangle (7 in this case) as the coefficients outside the cosines
3) Start with a cosine with a frequency of the same power, 7 in this case and just subtract 2 on each successive one until you either reach cos(1x) or cos(0x).

So another example for clarity:

\[\large \cos^6( \theta)=\frac{1}{2^5}[1\cos(6 \theta)+6\cos(4 \theta)+15\cos(2 \theta)+20 ]\]

In fact there is a similar (although not exactly the same) procedure for powers of sine and if anyone is interested I can show that method.

Additionally this is even more general if you know how to simply multiply matrices you can easily find out the general case\[\Large \sin^n \theta \cos^m \theta\] will be in terms of a simple sum of sines or cosines of different frequencies and amplitudes.