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Trigonometry Trigonometric Identity Questions

suzy123

New member
Dec 6, 2013
1
Your help will be greatly appreciated!

Thanks!


1. The expression \(\sin\pi\) is equal to \(0\), while the expression $\frac{1}{\csc\pi}$ is undefined. Why is $\sin\theta=\frac{1}{\csc\theta}$ still an identity?

2. Prove $\cos(\theta + \frac{\pi}{2})= -\sin\theta$
 
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SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
2. Prove $\cos(\theta + \frac{\pi}{2})= -\sin\theta$
Use the addition formula together with the values from the unit circle \(\displaystyle \cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Welcome to MHB, suzy123! :)

Your help will be greatly appreciated!

Thanks!


1. The expression \(\sin\pi\) is equal to \(0\), while the expression $\frac{1}{\csc\pi}$ is undefined. Why is $\sin\theta=\frac{1}{\csc\theta}$ still an identity?
It's an identity when both are defined.
And if you treat $\csc\pi$ as $\infty$ it even holds there.
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Welcome to MHB, suzy123! :)



It's an identity when both are defined.
And if you treat $\csc\pi$ as $\infty$ it even holds there.


An excellent point. Also, another way of looking at both is by considering limiting values:


\(\displaystyle \lim_{x \to \pi}\sin x=0\)

\(\displaystyle \lim_{x \to \pi}\csc x = \lim_{x \to \pi}\frac{1}{\sin x}=\frac{1}{ \lim_{x \to \pi}\sin x } = \lim_{z \to 0 }\frac{1}{z} \to \frac{1}{0} \to \infty\)


Similarly, you could use the composite angle formula I Like Serena gave above,

\(\displaystyle \sin (x \pm y)= \sin x \cos y \pm \cos x \sin y\)

and consider the limits


\(\displaystyle \sin \pi = \lim_{x \to \pi}\sin x= \lim_{\epsilon \to 0}\sin (\pi \pm \epsilon) \to 0\)


and


\(\displaystyle \csc \pi = \lim_{x \to \pi}\csc x= \lim_{\epsilon \to 0} \frac{1}{\sin (\pi \pm \epsilon)} \to \infty\)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Prove that the function:

$f(\theta) = \sin\theta\csc\theta$

has removable discontinuities at $k\pi,\ k \in \Bbb Z$.

It is, by and large, problematic to simply say:

$\infty = \frac{1}{0}$ because such an assignment does not obey the algebraic rules of the real numbers (in particular, the cancellation law:

$ac = bc \implies a = b$ when $c \neq 0$

breaks down).

While it *is* possible to extend the real numbers in various ways to include the notion of infinity, it is usually preferable to phrase statements about infinity in ways that do not mention infinity itself such as:

instead of:

$\displaystyle \lim_{x \to a} f(x) = \infty$

we say:

for any $N > 0$, there is some $\delta > 0$ such that for all $0 < |x - a| < \delta$, we have $f(x) > N$.

This is a fancy way of saying: $f$ increases without bound near $a$. Note it does not mention infinity, nor does it say what (if any) value we should ascribe to $f(a)$.

As others have mentioned, the cosecant function is undefined at certain points. If one is asked to evaluate cosecant at such a point, one ought to politely refuse.
 

DreamWeaver

Well-known member
Sep 16, 2013
337