# Math Help - Need Help:Inverse Trigonmetry Ratios

1. ## Need Help:Inverse Trigonmetry Ratios

How do I do these trig ratio's, and what formulas is the easiest, thanks..

Find all the angles x that satisfy the following equations;0˚ ≤ X < 360˚ Show your work:

1)sin(x)=0.4478.........................x=________ __and___________
2)cot(x)=-2.112........................x=__________and______ _____
3)csc(x)=5.2145........................x=_________ _and___________
4)cos(x)=0.7884........................x=_________ _and___________
5)sec(x)=-1.4557........................x=__________and_____ ______
6)cos(x)=-0.2298........................x=__________and_____ ______
7)sin(x)=-1.00 ........................x=__________and___________
8)tan(x)=0.6678........................x=_________ _and___________
9)sin(x)=0.0000........................x=_________ _and___________
10)cot(x)=0.8897........................x=________ __and___________
11)csc(x)=-2.0021........................x=__________and_____ ______
12)sin(x)=-0.1124........................x=__________and_____ ______
13)cos(x)=0.0000........................x=________ __and___________
14)tan(x)=-3.000........................x=__________and______ _____
15)cos(x)=1.000........................x=_________ _and___________
16)sec(x)=2.4012........................x=________ __and___________

2. How dare you!
I am not a textbook. I do not supply answers, I explain them!
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I am not going to answer all questions.
Rather tell you how to do them.
You need to use the inverse-trigonometric functions.

In the first example,
$\sin x = .4478$
Use the inverse trigonometric function on the calculator,
$\boxed{\boxed{ \sin^{-1}}}$
Therefore you get,
$x=26.60^o$
The other value is in the second quadrant,
$x=153.39^o$.

Note: when you reach a problem like,
$\sec x = 2.4012$
Is the same as,
$\cos x = .416$
Cuz, secant is reciprocal of cosine.

3. thank you for your help but how do you get the second quadrant?
and for the 3 question, how do you find csc(x)=5.2145. Ive tried $
\boxed{\boxed{ \sin^{-1}}}
$
and i got error on my calculator

4. Originally Posted by bigstarz
thank you for your help but how do you get the second quadrant?
Use the following formula.
Let $x$ be in the first quadrant.
Then, $180-x$---> Second quadrant.
Then, $180+x$---> Third quadrant.
Then, $360-x$---> Fourth quadrant.

It gets tricker when after you use the inverse function and get a negative number. If you cannot do those, first do the ones that do not give negative number and if you still do not understand how to work with negatives post your complaint here and someone will answer.
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Because you first need to find the reciprocal, that is,
$\frac{1}{5.2145}\approx .192$
Then you use the $\boxed{\boxed{ \sin^{-1}}}$math]

5. Hello, bigstarz!

I'll do one of each function . . . and hope it helps.

Find all the angles $x$ that satisfy the equations: . $0^o \leq x < 360^o$

$1)\;\sin x \:=\:0.4478$

Enter: $\boxed{\sin^{-1}}\;0.4478\;\boxed{=}$ . . . and get: $26.60262161\:\approx\:26.6^o$

$\text{sine}$ is positive in Quadrants 1 and 2.
. . Therefore: . $x\:=\:26.6,\;153.4^o$

$4)\;\cos x\:=\:0.7884$

Enter: $\boxed{\cos^{-1}}\;0.7884\;\boxed{=}$ . . . and get: $37.96376059 \:\approx\:38.0^o$

$\text{cosine}$ is positive in Quadrants 1 and 4.
. . Therefore: . $x \:=\:38.0^o,\;322.0^o$

$8)\;\tan(x)\:=\:0.6678$

Enter: $\boxed{\tan^{-1}}\;0.6678\;\boxed{=}$ . . . and get: $33.73499916 \:\approx\:33.7^o$

$\text{tangent}$ is positive in Quadrants 1 and 3.
. . Therefore: . $x \:=\:33.7^o,\;213.7^o$

$10)\;\cot x \:=\:0.8897$

Since $\cot x \:=\:\frac{1}{\tan x}$, enter: $0.8897\;\boxed{x^{-1}}$ . . . and get: $1.123974373$

Then enter: $\boxed{\tan^{-1}}\;\boxed{\text{ans}}\;\boxed{=}$ . . . and get: $48.3405102\:\approx\:48.3^o$

$\text{cotangent}$ is positive in Quadrants 1 and 3.
. . Therefore: . $x \:=\:48.3^o,\;228.3^o$

$(5)\;\sec x \:= \:-1.4557$

Since $\sec x \:= \:\frac{1}{\cos x}$, enter: $-1.4557\;\boxed{x^{-1}}$ . . . and get: $-0.68695473$

Then enter: $\boxed{\tan^{-1}}\;\boxed{\text{ans}}\;\boxed{=}$ . . . and get: $133.3895302\:\approx\:133.4^o$

$\text{secant}$ is negative in Quadrants 2 and 3.
. . Therefore: . $x \:=\:133.4^o,\;226.6^o$

$3)\;\csc x \:= \:5.2145$

Since $\csc x \:=\:\frac{1}{\sin x}$, enter: $5.2145\;\boxed{x^{-1}}$ . . . and get: $0.191772941$

Then enter: $\boxed{\sin^{-1}}\;\boxed{\text{ans}}\;\boxed{=}$ . . . and get: $11.05626911\:\approx\:11.1^o$

$\text{cosecant}$ is positive in Quadrants 1 and 2.
. . Therefore: . $x \:= \:11.1^o,\;168.9^o$