sin(tan^-1(-7/3)) How do i evaluate this? answer has to be exact with no decimals.. calculators not allowed either..
If you know the tangent to be 7/-3, you know the sin: -3/(sqrt(49+9) right?
The tangent is negative in II or IV
so sin(tan^-1(-7/3)
= 7/√58 or -7/√58
To evaluate the expression sin(tan^(-1)(-7/3) without a calculator and obtain an exact answer, you can follow these steps:
1. Draw a right triangle with an angle θ that satisfies tan^(-1)(-7/3).
2. Since tan^(-1)(-7/3) represents the angle whose tangent is (-7/3), the opposite side (O) is -7, and the adjacent side (A) is 3. Remember, in a right triangle, the hypotenuse (H) is constant and can be any value you choose. Let's choose H = 3 to keep things simple.
So, our triangle looks like this:
```
|
| θ
3 |________
| /
| /
-7 | /
```
3. Now, we can use the Pythagorean theorem to find the length of the hypotenuse (H):
H^2 = O^2 + A^2
H^2 = (-7)^2 + 3^2
H^2 = 49 + 9
H^2 = 58
Since H is positive, we take the positive square root of both sides:
H = √58
4. With the triangle's sides determined, we can evaluate sin(θ) using the following formula:
sin(θ) = O / H
sin(θ) = -7 / √58
5. To simplify the expression, rationalize the denominator by multiplying both the numerator and denominator by the conjugate of √58, which is √58:
sin(θ) = (-7 / √58) * (√58 / √58)
sin(θ) = -7√58 / 58
Therefore, the exact value of sin(tan^(-1)(-7/3)) is -7√58 / 58.