find sin theta and cos theta if cot theta = 3/7 and sec theta is < 0

Well, well, well, looks like we have a "trig-tricky" situation here. Let me put on my math clown nose and help you out!

Since cot(theta) is 3/7, we can use the Pythagorean identity to find the values of sin(theta) and cos(theta). The Pythagorean identity states that cot^2(theta) + 1 = csc^2(theta).

So, let's substitute the given value of cot(theta) into the equation:

(3/7)^2 + 1 = csc^2(theta)

Simplifying this, we get:

9/49 + 1 = csc^2(theta)

49/49 + 1 = csc^2(theta)

50/49 = csc^2(theta)

Taking the square root of both sides, we have:

csc(theta) = ± √(50/49)

Now, since sec(theta) is less than 0, we know that cos(theta) is negative.

The reciprocal relationship between the cosecant (csc) and sine (sin) functions tells us that:

csc(theta) = 1/sin(theta)

So, if csc(theta) = √(50/49), then sin(theta) = 1/√(50/49) = √(49/50).

Finally, since cosine (cos) is the reciprocal of secant (sec), we know:

cos(theta) = -1/sec(theta) = -1/√(50/49) = -√(49/50).

So, the values of sin(theta) and cos(theta) are approximately √(49/50) and -√(49/50), respectively.

Hope that mathematical clowning around didn't make you dizzy!

To find sin theta and cos theta, we need to first determine the quadrant in which theta lies.

Given that sec theta is < 0, we know that cosine is negative in either the second or third quadrant.

Now, we have cot theta = 3/7. We know that cot(theta) = cos(theta)/sin(theta).

Since cot(theta) is positive (3/7 is positive), both cos(theta) and sin(theta) must have the same sign. Therefore, they are both negative in either the second or third quadrant.

Now, let's consider the second quadrant. In the second quadrant, cos(theta) is negative, which matches our given information (sec theta < 0).

Therefore, cos(theta) < 0.

Now, we can find sin(theta) using the Pythagorean identity: cos^2(theta) + sin^2(theta) = 1.

Since cos(theta) < 0, let's assume cos(theta) = -x (where x is a positive number).

Using the Pythagorean identity: (-x)^2 + sin^2(theta) = 1, simplifying this equation: x^2 + sin^2(theta) = 1.

Since cos(theta) = -x and we know that cos^2(theta) + sin^2(theta) = 1, we can substitute -x for cos(theta) in the equation above and solve for sin(theta).

(-x)^2 + sin^2(theta) = 1

x^2 + sin^2(theta) = 1

sin^2(theta) = 1 - x^2

sin(theta) = sqrt(1 - x^2)

To find the value of x, we can use the given information that cot(theta) = 3/7.

cot(theta) = cos(theta)/sin(theta)

3/7 = -x / sqrt(1 - x^2)

Let's cross-multiply:

3 * sqrt(1 - x^2) = -7x

Square both sides to get rid of the square root:

9(1 - x^2) = 49x^2

Expanding and simplifying:

9 - 9x^2 = 49x^2

Rearranging the terms:

58x^2 = 9

Dividing both sides by 58:

x^2 = 9/58

Taking the square root of both sides:

x = sqrt(9/58)

Since cos(theta) = -x,

cos(theta) = -sqrt(9/58)

sin(theta) = sqrt(1 - x^2)

sin(theta) = sqrt(1 - (9/58))

After evaluating the above expressions, we find that

cos(theta) ≈ -0.387

sin(theta) ≈ -0.921

To find the values of sine (sin) and cosine (cos) of theta, we need to use the given information that cot(theta) = 3/7 and sec(theta) < 0.

We can start by using the fundamental trigonometric identities:

1. cot(theta) = cos(theta)/sin(theta)
2. sec(theta) = 1/cos(theta)

Given that cot(theta) = 3/7, we can substitute it into the first equation:

3/7 = cos(theta)/sin(theta)

Multiply both sides of the equation by sin(theta):

3sin(theta) = 7cos(theta)

Now, let's use the second piece of information that sec(theta) < 0.
Recall that sec(theta) = 1/cos(theta). Since sec(theta) is negative, we know that cos(theta) must also be negative.

Now, let's analyze the equation we obtained earlier: 3sin(theta) = 7cos(theta).
Since cos(theta) is negative, we can multiply both sides of the equation by -1 to change the sign:

-3sin(theta) = -7cos(theta)

Now we have two equations:

-3sin(theta) = -7cos(theta)
cos(theta) < 0

We can solve this system of equations by recognizing that sin(theta) = -7 and cos(theta) = -3.

So, the values of sin(theta) and cos(theta) are -7 and -3, respectively.

cotθ = 7/3 = y/x

If secθ = r/x < 0, then x<0 and y<0, so θ is in QIII
In QIII, both sinθ and cosθ < 0
r^2 = x^2 + y^2, so r = √40
sinθ = y/r
cosθ = x/r
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