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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.

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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
You're first step should be to draw the angle on the coordinate plane.
@alanli123 Fatal Error :P It's sin² x = 1 - cos² x You used sin x = 1 - cos² x

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Other answers:

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?
Quadrant 2
And...? There's another.
oh quadrant 3
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?
Quadrant 3 right?
Correct. So let's draw our triangle in quadrant 3... |dw:1361494367065:dw|
Now, the cosine of the angle is -2/3... how do you figure that into your triangle?
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)
put -2 on x-axis line and 3 on the longest side
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)
is it square root of 5?
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? :)
should it be - square root of 5?
Yes. it should :P So far, is everything clear to you? :) |dw:1361494965439:dw|
yes it is
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?
- sqrt 5/3
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?
3/ -sqrt 5
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...
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?
tangent of the *angle sorry
sqrt 5/2
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.
2/sqrt 5 or 2sqrt 5/5
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
ok that makes sense. thank you very much!!!
No problem :)
Need to go now, anyway :D ---------------------- Terence out

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