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3 years ago

Find the value of tan theta if sin theta = 12/13 and theta is in quadrant 2

Mathematics

1 answer:

8090 [49]3 years ago

5 0

Answer:

tanΘ = - $\frac{12}{5}$

Step-by-step explanation:

Using the trigonometric identities

• sin²x + cos²x = 1, hence

cosx = ± √(1 - sin²x )

• tanx = $\frac{sinx}{cosx}$

given sinΘ = $\frac{12}{13}$ , then

cosΘ = ± $\sqrt{1-(12/13)^2}$

Since Θ is in the second quadrant where cosΘ < 0, then

cosΘ = - $\sqrt{1-\frac{144}{169} }$

= - $\sqrt{\frac{25}{169} }$ = - $\frac{5}{13}$

tanΘ = $\frac{\frac{12}{13} }{\frac{-5}{13} }$

= $\frac{12}{13}$ × - $\frac{13}{5}$ = - $\frac{12}{5}$

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For the function defined by f(t)=2-t, 0≤t<1, sketch 3 periods and find:

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The half-range sine series is the expansion for $f(t)$ with the assumption that $f(t)$ is considered to be an odd function over its full range, $-1$ . So for (a), you're essentially finding the full range expansion of the function

$f(t)=\begin{cases}2-t&\text{for }0\le t$

with period 2 so that $f(t)=f(t+2n)$ for $|t|$ and integers $n$ .

Now, since $f(t)$ is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

$f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L$

where

$b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt$

In this case, $L=1$ , so

$b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt$
$b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{4-2(-1)^n}{n\pi}$

The half-range sine series expansion for $f(t)$ is then

$f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t$

which can be further simplified by considering the even/odd cases of $n$ , but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming $f(t)$ is even/symmetric across its full range. In other words, you are finding the full range series expansion for

$f(t)=\begin{cases}2-t&\text{for }0\le t$

Now the sine series expansion vanishes, leaving you with

$f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L$

where

$a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt$

for $n\ge0$ . Again, $L=1$ . You should find that

$a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3$

$a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt$
$a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}$
$a_n=\dfrac{2-2(-1)^n}{n^2\pi^2}$

Here, splitting into even/odd cases actually reduces this further. Notice that when $n$ is even, the expression above simplifies to

$a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0$

while for odd $n$ , you have

$a_{n=2k-1}=\dfrac{2-2(-1)^{2k-1}}{(2k-1)^2\pi^2}=\dfrac4{(2k-1)^2\pi^2}$

So the half-range cosine series expansion would be

$f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}a_{2k-1}\cos(2k-1)\pi t$
$f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t$

Attached are plots of the first few terms of each series overlaid onto plots of $f(t)$ . In the half-range sine series (right), I use $n=10$ terms, and in the half-range cosine series (left), I use $k=2$ or $n=2(2)-1=3$ terms. (It's a bit more difficult to distinguish $f(t)$ from the latter because the cosine series converges so much faster.)