trying to evaluate tan-1 (o) = -5.67

[davey2015]

hello guys

im trying to evaluate

tan-1 (o) = -5.67 (o) = degrees

what i have so far is

tan-1 (o) = -5.67
(o) = tan(-5.67)
tan(-5.67) = 0.09928

any help on this please as i havent done this before and tryed to learn at the same time.

thank you

 

[stapel]

im trying to evaluate

tan-1 (o) = -5.67 (o) = degrees

From what you do later, I think the exercise was something along the lines of the following:

. . . . .\(\displaystyle \mbox{Solve }\, \tan^{-1}(x)\, =\, -5.67,\, \mbox{ with }\, x\, \mbox{ in degrees.}\)

Please reply with confirmation or correction(s). Thank you!

 

[davey2015]

yeah we need to work out the angle. and work out/evaluate all other angles between 0 and 360 degrees.


cos−1(θ)=0.454
and
tan−1(θ)=-5.67

 

[Deleted member 4993]

yeah we need to work out the angle. and work out/evaluate all other angles between 0 and 360 degrees.

cos−1(θ)=0.454 ................ Θ here is not same
and
tan−1(θ)=-5.67 ................ as Θ here!

Are those statements parts of the same problem?

 

[stapel]

yeah we need to work out the angle. and work out/evaluate all other angles between 0 and 360 degrees.

cos−1(θ)=0.454
and
tan−1(θ)=-5.67

I will guess that the cosine equation is a different exercise, and that the "-1" in each equation is meant to be the exponent "^-1".

What did you get when you plugged this information into your calculator? What then did you do with what you know about the tangent curve over the stated interval? Where did you get stuck?

Please be complete. Thank you!

 

[Ishuda]

hello guys

im trying to evaluate

tan-1 (o) = -5.67 (o) = degrees

what i have so far is

tan-1 (o) = -5.67
(o) = tan(-5.67)
tan(-5.67) = 0.09928

any help on this please as i havent done this before and tryed to learn at the same time.

thank you

When you have the formula
tan^-1(a) = b
with no other information, the standard meaning is that a is a number with no units and b is an angle in radians [EDIT: That is b is the arctangent of a or, to put it another way, the tangent of b is a]. This can be changed to b is in degrees if it is so stated, in which case if you use b in an actual calculation you would need to convert b to radians. However, I have never seen the formula where the units of a is supposed to be degrees. Are you sure you have stated the problem correctly?

 

[davey2015]

here is a picture of the questions in hand. any help im really struggling here.

 

[stapel]

here is a picture of the questions in hand.

Okay; so we had correctly deciphered the intended text of your previous posts. Now, what have you done? You were provided with steps, but have demonstrated no effort to apply them. Are they not relevant?

I ask because, as formatted, the equations in the book make no sense. Inverse-trig functions take numbers as inputs; the angle-measures are their outputs. But, for some reason, your book has you doing something weird and completely non-standard with these equations. What did your book demonstrate when it showed worked examples for this type of "equation"?

Please provide all the steps that the book showed. When you reply, please also include a clear listing of everything you have done in your effort to mimic what the book outlined. Thank you!

 

[Deleted member 4993]

here is a picture of the questions in hand. any help im really struggling here.

Are you allowed to use calculator?