anonymous
  • anonymous
Find six trig values if sec(x)=-3/2 and tan(x)>0 I got cos(x)=-2/3 but that is all I could get.
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
if sec x = - 3/2 then cos x = -2/3 therefore sin x = 1 - (-2/3)^2 = 1- 4/9 = 5/9 therefore, csc x = 9/5 furthermore tan x = sin x / cos x = 5/9 / -2/3 = -5/6. however, since tan x > 0 we have that angle x must be in quadrant 1 or 3. since cos x is negative, the angle x must be in quadrant 3 and thus tan x = 5/6 (tan is positive in quadrant 3) now, finally cot x = 6/5
NoelGreco
  • NoelGreco
You're first step should be to draw the angle on the coordinate plane.
terenzreignz
  • terenzreignz
@alanli123 Fatal Error :P It's sin² x = 1 - cos² x You used sin x = 1 - cos² x

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

terenzreignz
  • terenzreignz
Hey, @haleyking345 Don't start writing down alanli's answers yet, they were all based on the conclusion that sin x = 5/9 which is WRONG Now relax and sit back... let's draw it on the coordinate plane, as @NoelGreco suggested. So, you know cosine is negative, and in which of the quadrants is cosine negative?
anonymous
  • anonymous
Quadrant 2
terenzreignz
  • terenzreignz
And...? There's another.
anonymous
  • anonymous
oh quadrant 3
terenzreignz
  • terenzreignz
Yep, Quadrant 2 or Quadrant 3. But it ALSO says that tangent is positive, and between quadrant 2 and quadrant 3, in only one of them is tangent positive, which is it?
anonymous
  • anonymous
Quadrant 3 right?
terenzreignz
  • terenzreignz
Correct. So let's draw our triangle in quadrant 3... |dw:1361494367065:dw|
terenzreignz
  • terenzreignz
Now, the cosine of the angle is -2/3... how do you figure that into your triangle?
terenzreignz
  • terenzreignz
Remember, cosine is the adjacent side over the hypotenuse... or x/r, where x is the length of the horizontal leg, and r is the radius (hypotenuse)
anonymous
  • anonymous
put -2 on x-axis line and 3 on the longest side
terenzreignz
  • terenzreignz
Very good. Hopefully, you figured out to put -2 on the x-axis line because it's to the left of the origin (0, 0) Anyway, |dw:1361494618728:dw| Now what's the length of the missing side? (HINT: Pythagorean Theorem)
anonymous
  • anonymous
is it square root of 5?
terenzreignz
  • terenzreignz
Almost... but look, this is the y-value, the vertical leg of the triangle, right? And it's in the third Quadrant (QIII) And in the third quadrant, both x and y should be negative, no? :)
anonymous
  • anonymous
should it be - square root of 5?
terenzreignz
  • terenzreignz
Yes. it should :P So far, is everything clear to you? :) |dw:1361494965439:dw|
anonymous
  • anonymous
yes it is
terenzreignz
  • terenzreignz
Okay, so now, you can easily determine the sine of the angle :) sine is the opposite side over the hypotenuse, or x/r What's the sine?
anonymous
  • anonymous
- sqrt 5/3
terenzreignz
  • terenzreignz
I saw the sine...and it opened up my LOL Yes, correct! And with the sine, you can easily get the cosecant :D What is it?
anonymous
  • anonymous
3/ -sqrt 5
terenzreignz
  • terenzreignz
Yeah... but some teachers can be picky and make you rationalise the denominator... fine... \[\huge -\frac{3}{\sqrt{5}}=-\frac{3\sqrt{5}}{5} \] May as well put it up like that...
terenzreignz
  • terenzreignz
Now, back to the triangle. You can also now easily determine the tangent of the triangle :) Tangent is the opposite side, over the adjacent side or y/x What's the tangent?
terenzreignz
  • terenzreignz
tangent of the *angle sorry
anonymous
  • anonymous
sqrt 5/2
terenzreignz
  • terenzreignz
Yep :) As you can see, the tangent is positive. Now with the tangent, you can easily get the cotangent. What is it? Also, rationalise it.
anonymous
  • anonymous
2/sqrt 5 or 2sqrt 5/5
terenzreignz
  • terenzreignz
And there you have it, your six trigonometric functions :D I'd like to point out that had alanli used the correct identity for sine, that is sin²x = 1 - cos²x it would have given... sin²x = 1 - (-2/3)² = 1 - (4/9) = 5/9 But you still have to get the square root of this sin x = (+/-) sqrt(5)/sqrt(9) = (+/-)sqrt(5)/3
anonymous
  • anonymous
ok that makes sense. thank you very much!!!
terenzreignz
  • terenzreignz
No problem :)
terenzreignz
  • terenzreignz
Need to go now, anyway :D ---------------------- Terence out

Looking for something else?

Not the answer you are looking for? Search for more explanations.