cotx = -3/5,-pi/2<x<0
Use identities to find the exact value for the remaining five trigonometric functions
???? How can you say that cotx = -3/4, when you started out given that cotx = -3/5?
Draw a triangle in QIV with
x = 3
y = -5
so r = √34
sinθ = y/r = -5/√34
cosθ = x/r = 3/√34
tanθ = y/x = -5/3
cotθ = x/y = -3/5
secθ = r/x = √34/3
cscθ = r/y = -√34/5
To find the exact values of the remaining five trigonometric functions (sin, cos, tan, sec, and csc) given that cot(x) = -3/5 and -pi/2 < x < 0, we can use the following trigonometric identities:
1. sin^2(x) + cos^2(x) = 1
2. tan(x) = sin(x)/cos(x)
3. sec(x) = 1/cos(x)
4. csc(x) = 1/sin(x)
Let's use these identities step by step:
Step 1: Start with the given information: cot(x) = -3/5.
Step 2: Recall that cot(x) is the reciprocal of tan(x). Therefore, tan(x) = -5/3.
Step 3: Recall that tan(x) is the ratio of sin(x) to cos(x). We can set up the equation as follows:
-5/3 = sin(x) / cos(x).
Step 4: Given that -pi/2 < x < 0, we know that x lies in the fourth quadrant of the unit circle (where sin(x) < 0 and cos(x) < 0).
Step 5: Since sin(x) < 0 and cos(x) < 0, we can assign positive values to sin(x) and cos(x) by taking the absolute values.
Step 6: Let's assume that sin(x) = a and cos(x) = b, where a is positive and b is negative.
Step 7: Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we substitute the values of a and b into this equation:
a^2 + b^2 = 1.
Step 8: From Step 7, we can solve for a^2:
a^2 = 1 - b^2.
Step 9: Setting the equation from Step 3 into the equation from Step 8, we have:
-5/3 = a / b.
Step 10: Squaring both sides of this equation, we get:
25/9 = a^2 / b^2.
Step 11: Using the equation from Step 8, we can substitute for a^2:
25/9 = (1 - b^2) / b^2.
Step 12: Simplifying this equation, we have:
25b^2 = 9 - 9b^2.
Step 13: Move all terms to one side of the equation:
34b^2 = 9.
Step 14: Divide both sides by 34 to solve for b^2:
b^2 = 9/34.
Step 15: Take the square root of both sides to solve for b:
b = ±√(9/34) = ±3/√34.
Step 16: Since we assigned b as negative, we take b = -3/√34.
Step 17: Substitute this value of b into a^2 = 1 - b^2 to solve for a:
a^2 = 1 - (-3/√34)^2.
Step 18: Simplify and solve for a^2:
a^2 = 1 - 9/34 = 25/34.
Step 19: Take the square root of both sides to solve for a:
a = ±√(25/34) = ±5/√34.
Step 20: Since we assigned a as negative, we take a = -5/√34.
Based on the steps above, the exact values for the trigonometric functions are:
sin(x) = a = -5/√34
cos(x) = b = -3/√34
tan(x) = sin(x) / cos(x) = (-5/√34) / (-3/√34) = 5/3
sec(x) = 1 / cos(x) = 1 / (-3/√34) = -√34 / 3
csc(x) = 1 / sin(x) = 1 / (-5/√34) = -√34 / 5
So, the exact values for the remaining five trigonometric functions are:
sin(x) = -5/√34
cos(x) = -3/√34
tan(x) = 5/3
sec(x) = -√34 / 3
csc(x) = -√34 / 5
We know that cotx = -3/5.
Since cotx = cosx/sinx, we can solve for sinx and cosx using the Pythagorean identity, sin^2x + cos^2x = 1:
cotx = cosx/sinx
-3/5 = cosx/sinx
Squaring both sides:
9/25 = cos^2x/sin^2x
Using the Pythagorean identity, we know that sin^2x + cos^2x = 1, so we can substitute cos^2x for 1 - sin^2x:
9/25 = (1 - sin^2x)/sin^2x
Multiplying both sides by sin^2x:
9 = 25 - 25sin^2x
25sin^2x = 16
sinx = +/- 4/5
If sinx = 4/5, then cosx = -3/5 (since cotx is negative in the given interval).
Now, we can find the remaining trigonometric functions:
tanx = sinx/cosx = (4/5)/(-3/5) = -4/3
cscx = 1/sinx = 5/4
secx = 1/cosx = -5/3
cotx = cosx/sinx = (-3/5)/(4/5) = -3/4
Therefore, the exact values of the six trigonometric functions are:
sinx = 4/5
cosx = -3/5
tanx = -4/3
cscx = 5/4
secx = -5/3
cotx = -3/4
I apologize for the mistake in my previous response. As you correctly pointed out, cotx = -3/5, and not -3/4. Here are the correct values for the remaining trigonometric functions:
sinx = -4/5 (since sinx is negative in the given interval)
cosx = -3/5
tanx = sinx/cosx = (-4/5)/(-3/5) = 4/3
cscx = 1/sinx = -5/4
secx = 1/cosx = -5/3
cotx = cosx/sinx = (-3/5)/(-4/5) = 3/4
Again, I apologize for any confusion my previous response may have caused. I appreciate you pointing out the error.