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anonymous
 3 years ago
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
 3 years ago
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.

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0if 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
 3 years ago
Best ResponseYou've already chosen the best response.0You're first step should be to draw the angle on the coordinate plane.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0@alanli123 Fatal Error :P It's sin² x = 1  cos² x You used sin x = 1  cos² x

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Hey, @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?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0And...? There's another.

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Yep, 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?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Correct. So let's draw our triangle in quadrant 3... dw:1361494367065:dw

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Now, the cosine of the angle is 2/3... how do you figure that into your triangle?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Remember, 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
 3 years ago
Best ResponseYou've already chosen the best response.0put 2 on xaxis line and 3 on the longest side

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Very good. Hopefully, you figured out to put 2 on the xaxis 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
 3 years ago
Best ResponseYou've already chosen the best response.0is it square root of 5?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Almost... but look, this is the yvalue, 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
 3 years ago
Best ResponseYou've already chosen the best response.0should it be  square root of 5?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Yes. it should :P So far, is everything clear to you? :) dw:1361494965439:dw

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, 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?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0I 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?

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah... 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
 3 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 3 years ago
Best ResponseYou've already chosen the best response.0tangent of the *angle sorry

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Yep :) 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
 3 years ago
Best ResponseYou've already chosen the best response.02/sqrt 5 or 2sqrt 5/5

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0And 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok that makes sense. thank you very much!!!

terenzreignz
 3 years ago
Best ResponseYou've already chosen the best response.0Need to go now, anyway :D  Terence out
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