given cscx= -3 and tanx>0 find cosx
First identify the quadrant in which the angle lies. Since tanx>0 (posotive) it can be in either Quadrant I or III. Because sin is negative, it must be in quadrant III.
To reach familiar terms, change cscx= -3 to sinx= -1/3 .
Make a triangle and use sin= -1/3 to find the missing side.
-----missing side= sqrt(8) or 2sqrt(2)
therefore, cosx= - [2sqrt(2)]/3
Remeber that cos is negative in Quadrant III.
To find the value of cos(x) given that csc(x) = -3 and tan(x) > 0, we can use the relationship between trigonometric functions.
We are given that csc(x) = -3. The reciprocal of csc(x) is sin(x), which means sin(x) = -1/3. Now, using the Pythagorean identity, we can find cos(x).
The Pythagorean identity states that sin^2(x) + cos^2(x) = 1.
Substituting sin(x) = -1/3, we have (-1/3)^2 + cos^2(x) = 1.
Simplifying this equation, we get 1/9 + cos^2(x) = 1.
Rearranging the equation, we get cos^2(x) = 8/9.
Since tan(x) = sin(x) / cos(x), and tan(x) > 0, we know that sin(x) and cos(x) have the same sign. Since sin(x) = -1/3, cos(x) must be negative. Therefore, cos(x) = -√(8/9) = -2√2/3.
So, the value of cos(x) is -2√2/3.