Trigonometric functions

[rachelmaddie]

Hi, I need my work checked for this please.

The expected values of the five remaining trigonometric functions of θ are:

Tangent
tan θ = 1 / cot (θ) = 1 /[-6/7] = -7/6

Secant
sec^2θ = 1 + tan^2θ = 1 + (-7/6)^2 = 1 + 49/36 = 85/36
sec θ = ± (√85)/ 6
Secant is positive in Quadrant four.
sec θ = (√85)/ 6

Cosine
cosθ = 1 / secθ = 6 / (√85 = 6
(√85) / 85

Sine
sin^2θ + cos^2θ = 1 —> sin^2θ = 1 - cos^2θ = 1
-[6(√85)/85]^2 = sin^2θ = 1 - 36*85/(85)^2 = 1- 36/85 = 49/85
sinθ = ± 7 / (√85) = ± 7 (√85) / 85
Sine is negative in Quadrant four.
sinθ = - 7 (√85)/ 85

Cosecant
cscθ = 1 / sinθ = - 85 / (7√85) = (- √85) / 7

These are the five remaining trigonometric functions:
tanθ = - 7/6
secθ = (√85)/ 6
cosθ = 6 (√85)/ 85
sinθ = - 7 (√85)/ 85
cscθ = - (√85)/ 7

 

[Deleted member 4993]

Hi, I need my work checked for this please.
View attachment 20657
The expected values of the five remaining trigonometric functions of θ are:

Tangent
tan θ = 1 / cot (θ) = 1 /[-6/7] = -7/6

Secant
sec^2θ = 1 + tan^2θ = 1 + (-7/6)^2 = 1 + 49/36 = 85/36
sec θ = ± (√85)/ 6
Secant is positive in Quadrant four.
sec θ = (√85)/ 6

Cosine
cosθ = 1 / secθ = 6 / (√85 = 6
(√85) / 85

Sine
sin^2θ + cos^2θ = 1 —> sin^2θ = 1 - cos^2θ = 1
-[6(√85)/85]^2 = sin^2θ = 1 - 36*85/(85)^2 = 1- 36/85 = 49/85
sinθ = ± 7 / (√85) = ± 7 (√85) / 85
Sine is negative in Quadrant four.
sinθ = - 7 (√85)/ 85

Cosecant
cscθ = 1 / sinθ = - 85 / (7√85) = (- √85) / 7

These are the five remaining trigonometric functions:
tanθ = - 7/6
secθ = (√85)/ 6
cosθ = 6 (√85)/ 85
sinθ = - 7 (√85)/ 85
cscθ = - (√85)/ 7

Use your calculator to check.

What do you find?

 

[rachelmaddie]

Use your calculator to check.

What do you find?

I’m getting different values?

 

[Otis]

I’m getting different values?

Hi rachel. I'm not sure what you did with the calculator, but your answers in post #1 look good to me. (I didn't check your work; I drew the reference angle -- to label sides of a right-triangle -- and then used the right-triangle trig definitions.)

?

 

[rachelmaddie]

Hi rachel. I'm not sure what you did with the calculator, but your answers in post #1 look good to me. (I didn't check your work; I drew the reference angle -- to label sides of a right-triangle -- and then used the right-triangle trig definitions.)

?

Can you check my work please?

 

[pka]

View attachment 20657
The expected values of the five remaining trigonometric functions of θ are:

To rachelmaddie, do you know, I mean really know, as in you have no need to look these up?
\(\Large{\tan(\theta)=\dfrac{y}{x},~\sin(\theta)=\dfrac{y}{r},~\&~\cos(\theta)=\dfrac{x}{r}}\) Where \(\large r=\sqrt{x^2+y^2}\)
rachelmaddie I really me that you should take tine to memorize those basic facts, along with the signs of \(x~\&~y\) in the various quadrants.

Lets apply those to the question at hand: \(\cot(\theta)=-\dfrac{6}{7}\) where \(\theta\in IV\).
Using that given and what you know, \(\large x=6,~y=-7,~\&~r=\sqrt{85}\)
If you have learned the basics it is easy, easy to complete the list:
\(\sin(\theta)=\dfrac{y}{r}=\dfrac{-7}{\sqrt{85}}\\\cos(\theta)=\dfrac{x}{r}=\dfrac{6}{\sqrt{85}}\\\tan(\theta)=\dfrac{y}{x}=\dfrac{-7}{6}\\ \sec(\theta)=\dfrac{r}{x}=\dfrac{\sqrt{85}}{6}\\\csc(\theta)=\dfrac{r}{y}=\dfrac{\sqrt{85}}{-7}\)

You Too can do that by simply learning the simple basics.

 

[Deleted member 4993]

I’m getting different values?

Did you first calculate the value of Θ, using calculator, from tan (Θ) = -7/6 ?

What value did you get?

 

[pka]

Did you first calculate the value of Θ, using calculator, from tan (Θ) = -7/6 ?
What value did you get?

The question does not ask for the value \(\theta\).

 

[Dr.Peterson]

The question does not ask for the value \(\theta\).

What's under discussion now is post #3, saying that a check using the angles gave different values than the previous correct work without finding angles (in post #1).