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.

- 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.

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- 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

You're first step should be to draw the angle on the coordinate plane.

- terenzreignz

@alanli123
Fatal Error :P
It's
sin² x = 1 - cos² x
You used
sin x = 1 - cos² x

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- 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

Quadrant 2

- terenzreignz

And...? There's another.

- anonymous

oh quadrant 3

- 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

Quadrant 3 right?

- terenzreignz

Correct. So let's draw our triangle in quadrant 3...
|dw:1361494367065:dw|

- terenzreignz

Now, the cosine of the angle is -2/3... how do you figure that into your triangle?

- 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

put -2 on x-axis line and 3 on the longest side

- 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

is it square root of 5?

- 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

should it be - square root of 5?

- terenzreignz

Yes. it should :P
So far, is everything clear to you? :)
|dw:1361494965439:dw|

- anonymous

yes it is

- 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

- sqrt 5/3

- 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

3/ -sqrt 5

- 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

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

tangent of the *angle
sorry

- anonymous

sqrt 5/2

- 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

2/sqrt 5 or 2sqrt 5/5

- 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

ok that makes sense. thank you very much!!!

- terenzreignz

No problem :)

- terenzreignz

Need to go now, anyway :D
----------------------
Terence out

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