trig (1 Viewer)

turnerloos · Oct 9, 2015

can someone please explain why sin3x = cos (3x - π/2)?

InteGrand · Oct 9, 2015

turnerloos said:
can someone please explain why sin3x = cos (3x - π/2)?

$The following is a trigonometric identity: $\sin \theta = \cos \left(\theta - \frac{\pi}{2} \right)$.$

$\textbf{Proof.}$

$$\sin \theta = \cos \left(\frac{\pi}{2}-\theta \right)$ (complementary angles trig. identity)$

$$= \cos \left[ - \left( \frac{\pi}{2}-\theta\right) \right]$ (cosine function is even)$

$= \cos \left(\theta -\frac{\pi}{2} \right)$. $\blacksquare$

Flop21 · Oct 9, 2015

Make 3x = theta and it becomes more clear. pi/2 is 90 degrees and it's just something you know, cos(theta - 90) = sin theta.

InteGrand · Oct 9, 2015

Flop21 said:
it's just something you know, cos(theta - 90) = sin theta.

$I never memorised this personally. If this result was ever needed, I'd just picture the graph of $\cos \theta$, shift it to the right by $\frac{\pi}{2}$ to get the graph of $\cos \left(\theta -\frac{\pi}{2} \right)$, and note that this shifted graph coincides with that of $\sin \theta$, so $\cos \left(\theta -\frac{\pi}{2} \right) = \sin \theta$ for all real $\theta$. Alternatively, derive via proof given above.$

turnerloos · Oct 9, 2015

so would sin 3x = cos (π/2 - 3x) be right as well?

InteGrand · Oct 9, 2015

turnerloos said:
so would sin 3x = cos (π/2 - 3x) be right as well?

$Yes, that's just the complementary angles trig. identity: $\sin \alpha = \cos \left(\frac{\pi}{2}-\alpha \right)$ for any number $\alpha$.$

leehuan · Oct 9, 2015

Flop21 said:
Make 3x = theta and it becomes more clear. pi/2 is 90 degrees and it's just something you know, cos(theta - 90) = sin theta.

Not really... cos(90-x)=sin(x) is the important one to memorise.

EDIT:
Other one being cos(-x)=cos(x)

Flop21 · Oct 9, 2015

leehuan said:
Not really... cos(90-x)=sin(x) is the important one to memorise.

I don't understand, is what I said wrong?

How do we derive these identities?? If anyone has time to explain it to me I would appreciate it. The graph way integrand mentioned is a good idea but is there another way?

InteGrand · Oct 9, 2015

Flop21 said:
I don't understand, is what I said wrong?

How do we derive these identities?? If anyone has time to explain it to me I would appreciate it. The graph way integrand mentioned is a good idea but is there another way?

$Nothing you said was mathematically wrong, it's just that you don't \textit{need} to memorise the identity you said, because it is quickly derived from the typical one $\sin x = \cos \left( \frac{\pi}{2}-x\right)$ by using the fact that $\cos$ is even. Since you already know (or should know) these two facts, no point memorising the one you mentioned, since you probably have enough to memorise as it is. Just learn the basic facts like the complementary angles identities and supplementary angles identities, the functions' periodicities, and whether the functions are odd or even, and you can derive the other identities from these.$

InteGrand · Oct 9, 2015

Flop21 said:
I don't understand, is what I said wrong?

How do we derive these identities?? If anyone has time to explain it to me I would appreciate it. The graph way integrand mentioned is a good idea but is there another way?

$Yes, I gave another way to derive this one near the top of this page. Basically $\sin x = \cos \left( \frac{\pi}{2}-x\right) $ by complementary angle trig. identity, and this is just $\cos \left( x - \frac{\pi}{2}\right)$ since the cosine function is even.$

InteGrand · Oct 9, 2015

$Basically, these are the key \textsl{basic} trig. identities you need to know or are helpful to know to be able to derive the others (not going in to double angle and compound angles identities here, or complicated identities):$

$\underline{Odd/even properties}$

$$\bullet$ $\sin (-x) = -\sin x$ (sine function is odd)$

$$\bullet$ $\cos (-x) = \cos x$ (cosine function is even)$

$$\bullet$ $\tan (-x) = -\tan x$ (tangent function is odd)$

$\underline{Periodicity}$

$\bullet$ Period of sine is $2\pi$: $\sin x =\sin (x+2n\pi)$ for any integer $n$

$\bullet$ Period of cosine is $2\pi$: $\cos x =\cos (x+2n\pi)$ for any integer $n$

$\bullet$ Period of tangent is $\pi$: $\tan x =\tan (x+n\pi)$ for any integer $n$

$\underline{Complementary angle identities}$

$\bullet$ $\sin x = \cos \left( \frac{\pi}{2}-x\right)$

$\bullet$ $\cos x = \sin \left( \frac{\pi}{2}-x\right)$

$\bullet$ $\tan x = \cot \left( \frac{\pi}{2}-x\right)$

$\underline{Supplementary angle identities}$

$\bullet$ $\sin x = \sin \left( \pi-x\right)$

$\bullet$ $\cos x = -\cos \left( \pi-x\right)$

$\bullet$ $\tan x = -\tan \left( \pi-x\right)$

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trig (1 Viewer)

turnerloos

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