Trig Functions Help... I'm not even sure what to do.

given the instructions to __find the following if i am given that sinx=3/5 while cosx<0 and tanr=-8/15 while sinr>0 __(x and r being general variables...)

a. cosx

b. sec (r)

c. sin (x-pi)

d. tan (-r)

I have absolutley no idea how to go about this problem and it would be very helpful is someone could help me with just one of the problems so i understand the process... what exactly is it asking?

Thank you

How would i continue that process for problems such as sin(x-pi)? I don't remember ever using triangles for these problems... only algebra... so i'm a bit confused

Re: Trig Functions Help... I'm not even sure what to do.

Quote:

Originally Posted by

**FxAperture** given the instructions to __find the following if i am given that sinx=3/5 while cosx<0 and tanr=-8/15 while sinr>0 __(x and r being general variables...)

a. cosx

b. sec (r)

c. sin (x-pi)

d. tan (-r)

I have absolutley no idea how to go about this problem and it would be very helpful is someone could help me with just one of the problems so i understand the process... what exactly is it asking?

Thank you

sin(x) = 3/5, being a positive value, must be in the first or second quadrants, with cos(x) < 0 as well meaning that it is in the second quadrant. Draw a right angle triangle with opposite side 3 units and hypotenuse 5 units, use Pythagoras to find the adjacent side (if you can't already tell it's a 3,4,5 triangle), state what cos(x) is, and note that you need to negate this value because you're in the second quadrant.

Follow similar procedures for the rest.

Re: Trig Functions Help... I'm not even sure what to do.

It might help to draw a graph of a circle centered at (0,0). The radius of the circle is the hypotenuse of the right triangle (so the length of the radius is the length of the hypotenuse). The radius has endpoints (0,0) (i.e. the origin) and (x,y) (i.e. where the radius meets the circumference of the circle). x is your "adjacent" value, and "y" is your "opposite" value. The angle that you want is from the positive x-axis to the radius. For part c), it might help to recall that pi is 180 degrees. To draw the triangle, draw a vertical line to the x-axis from the appropriate endpoint of the radius (the appropriate endpoint is where the radius meets the circumference of the circle)

You'll have to be careful because there can be more than one triangle that represents your trig function. For example, the sin(x)=3/5 can be satisfied by two triangles. One in the first quadrant (an endpoint of the radius is (4,3)), and one in the second quadrant (an endpoint of the radius is (-4,3)). If you look at those two triangles, they satisfy sin(x)=3/5 (both have radius 5). You're given that cos(x)<0, so that means the x value should be negative, so we want the triangle in the second quadrant, with corresponding endpoint (-4,3). So if you follow what I outlined above, "adjacent" has value "-4", hypotenuse is 5, so cos(x) is -4/5.

So, what you basically do is find all the triangles (usually 2) that satisfy the given trig function (i.e. sin(x)=3/5 or tan(r)=-8/15), then find which of those is the triangle that satisfies the condition (i.e. cos(x)<0 or sin(r)>0). Once you have your triangles, use the appropriate endpoint of the radius to determine what your "adjacent", "opposite", and "hypotenuse" values are in order to compute the problem.

For part c, recall that pi is 180 degrees ... so if your sin(x)="something", then sin(x-pi) would be sin corresponding to the radius of the original (i.e. "something") adjusted by moving it back 180 degrees.

As for the possible triangles of tan, remember that a/b = (-a) / (-b), and -a/b = (-a) / b = a / (-b), where your a and b are the x and y coordinates which can be negative. This idea doesn't really work for sin and cos, but remember that the hypotenuse is the length of the radius, which will always be positive. For sin and cos, the idea is that if sin depends on the y value and radius, so the x value can be positive or negative (you can get the x value by using a^2+b^2=r^2).