Trigonmetry Functions (radian circles)

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Given sin(x) = -6/7 in the third quadrant, the cosine and tangent values can be derived using the Pythagorean theorem. The adjacent side can be calculated as -√(7² - 6²), which results in a negative value due to the quadrant. Consequently, cos(x) is negative and can be expressed as the ratio of the adjacent side to the hypotenuse. Finally, tan(x) is found by dividing sin(x) by cos(x), leading to the necessary fractional forms for both trigonometric functions. Understanding these relationships allows for accurate calculations of cosine and tangent based on the given sine value.
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Homework Statement


If sin (x) = -6/7

what is cos(x) = ?
what is tan(x) = ?

the answers have to be in integer form, as in no decimals.
the "x" is supposed to be in quadrant 3

2. The attempt at a solution

This is in radians

so i did sin^-1 (-6/7) = -1.02969

the answer I am assuming is (x)

than cos (x) --> 0.5150 and since its in quadrant 3 this will -.5150
than tan(x) --> -1.664

im not sure if i am right soo far, but assuming i am. i tried using pythagoras to figure our what -0.5150 will be in decimal --> -(sqrt(13))/7

and tan --> and I am lost can't figure this out

any guide or assistance would be great thanks
 
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sin(x)=opp/hyp.

so if sin(x)=-6/7, what is opp and hyp?

When you get that, draw the angle x in the 3rd quadrant and construct the right angled triangle.
 
Draw a right triangle inside a circle of radius 7. The sine function is positive in quadrants I and II and is negative in quadrants III and IV. Since you are given that sin(x) = -6/7, which quadrants can the angle x be in? For each of these, how long does the side adjacent have to be?

Your answers won't be in integers: they will be either rational numbers (the quotient of two integers) or the quotient of a radical and an integer.
 
You could also use the fact that sin^2(x)+ cos^2(x)= 1 to find cos(x) (remembering that the angle is in the third quadrant so cos(x) is negative.)

Then use the fact that tan(x)= sin(x)/cos(x).

In essence, both rock.freak667 and I are telling you, "yes, use the Pythagorean theorem".
 
The teacher asked for fractions, so my mistake as putting it as an integer sorry..

I get how to get what tan(x) is but i don't understand how to find the fraction or what the fraction looks like

the cos one worked, ( i can check online if i got the right answer and i did) but tan is confusing me,
 
phat2107 said:
The teacher asked for fractions, so my mistake as putting it as an integer sorry..

I get how to get what tan(x) is but i don't understand how to find the fraction or what the fraction looks like

the cos one worked, ( i can check online if i got the right answer and i did) but tan is confusing me,

tan(x)=sin(x)/cos(x)

or if you drew the triangle properly, you could find tan(x) by it's definition.
 
Use the fact that:

cos(x)=\sqrt{1-sin^2x}

If you know the value of sin(x) you can very easy find the value for cos(x) using the formula above ^^.

Then, when you know the values for sin(x) and cos(x), just substitute here tan(x)=sin(x)/cos(x) and find tan(x).

Regards.
 
Rock.freak667's point is that, since sine= opp/hyp, you can represent sin(x)= -6/7 as a right triangle with "opp"= -6 (on a graph, down 6 on the y-axis) and hypotenuse 7. You can use the Pythagorean theorem to determine that "adj" (measured on the x-axis) is -\sqrt{7^2- 6^2} (negative square root because you are told that the angle is in the third quadrant and both x and y are negative there). Now that you know the lengths of all three sides of the right triangle, it is easy to find any of the trig functions. In particular, cos(x)= adj/hyp and tan(x)= opp/adj.
 

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