Question

1. # If Cos Θ = Square Root 2 Over 2 And 3 Pi Over 2 < Θ < 2Π, What Are The Values Of Sin Θ And Tan Θ?

Trigonometry is an essential part of mathematics, allowing us to use angles and lengths to analyze shapes and figures. It’s an important topic to understand in a variety of fields, from physics and engineering to navigation and surveying. This article will explore the problem posed above; if cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2Π, what are the values of sin Θ and tan Θ? We’ll look at how we can solve this problem with what we know about trigonometric functions. Read on to learn more!

## What is cos Θ?

The cosine of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse. In mathematical terms, it is written as: cos Θ = a/h. where a is the length of the adjacent side and h is the length of the hypotenuse. The value of cos Θ lies between -1 and 1.

## What is sin Θ?

The value of sin Θ can be found by using the cosine function. Cos Θ = Square Root Over And Pi Over < Θ < Π, so the value of sin Θ is 1 – (Square Root Over And Pi Over ). The value of tan Θ can be found by using the tangent function. Tan Θ = (1 – Square Root Over And Pi Over ) / (Square Root Over And Pi Over ), so the value of tan Θ is 1 + (1 – Square Root Over And Pi Over ).

## What is tan Θ?

tan Θ is the ratio of the length of the side opposite Θ to the length of the side adjacent to Θ.

## How to solve for sin Θ and tan Θ when given cos Θ

If you know the value of cos Θ, you can use a few simple steps to solve for sin Θ and tan Θ. First, recall the trigonometric identity that states:

cos^2(Θ) + sin^2(Θ) = 1

This identity is useful because it allows you to solve for either sine or cosine if you know the other. In our case, we know cos(Θ), so we can solve for sin(Θ). Rearranging the equation above, we get:

sin^2(Θ) = 1 – cos^2(Θ)

Now we can take the square root of both sides to solve for sin(Θ):

sin(Θ) = +/- sqrt[1 – cos^2(Θ)]

Since we are given that 0 < Θ < π, we know that sin(Θ) will be positive in this case. Therefore, our final equation for sin(Θ) is:

sin(Θ) = sqrt[1 – cos^2(Θ)]

## Conclusion

In conclusion, we have seen that given the values of cos θ and the range of θ, it is possible to calculate both sin θ and tan θ. We found that sinθ = 1/√2 and tanθ = √2 respectively. Although this may seem daunting at first, with a little bit of practice it can become easier to understand and apply these concepts. With this knowledge in hand you can go about solving problems involving trigonometric functions with ease!

2. If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2Π, what are the values of sin Θ and tan Θ?

It might seem like a tricky problem, but it’s actually quite simple if you remember a few mathematical principles. To start off, let’s define sin Θ and tan Θ. Sin Θ is the sine of an angle, which is a ratio between the length of the side opposite the angle and the length of the hypotenuse. Tan Θ is the tangent of an angle, which is a ratio between the length of the side opposite the angle and the length of the side adjacent to the angle.

Using the given information, we can use trigonometric identities to find the values of sin Θ and tan Θ. Since cos Θ is equal to the square root of 2 over 2, we can use the Pythagorean Theorem to find the lengths of the side opposite and adjacent to the angle. The length of the side opposite the angle is equal to the square root of 2, and the length of the side adjacent to the angle is equal to 1.

Now that we have the lengths of the sides, we can use them to calculate the values of sin Θ and tan Θ. Sin Θ is equal to the length of the side opposite the angle divided by the length of the hypotenuse, which is equal to square root of 2 over 2. Tan Θ is equal to the length of the side opposite the angle divided by the length of the side adjacent to the angle, which is equal to square root of 2 over 1.

Therefore, if cos Θ is equal to the square root of 2 over 2 and 3 pi over 2 < Θ < 2Π, the values of sin Θ and tan Θ are equal to square root of 2 over 2 and square root of 2 over 1, respectively.