We can start by using the Pythagorean identity for tangent:
$$\tan^2\theta + 1 = \sec^2\theta$$
Since we know $\tan\theta = 2$, we can find $\sec\theta$:
$$\tan^2\theta + 1 = \sec^2\theta$$
$$2^2 + 1 = \sec^2\theta$$
$$\sec\theta = \sqrt{5}$$
Next, we can use the identity $\sin^2\theta + \cos^2\theta = 1$ to find $\cos\theta$:
$$\sin^2\theta + \cos^2\theta = 1$$
$$\cos^2\theta = 1 - \sin^2\theta$$
We can express $\sin\theta$ in terms of $\cos\theta$ by using the identity for tangent:
$$\tan\theta = \frac{\sin\theta}{\cos\theta}$$
$$\sin\theta = \tan\theta\cos\theta$$
$$\sin\theta = 2\cos\theta$$
Substituting this into the equation above, we get:
$$\cos^2\theta = 1 - (2\cos\theta)^2 = 1 - 4\cos^2\theta$$
Solving for $\cos\theta$, we get:
$$\cos^2\theta = \frac{1}{5}$$
$$\cos\theta = \pm\frac{1}{\sqrt{5}}$$
Since $\tan\theta$ and $\sin\theta$ have the same sign (both positive in this case), we can take the positive value for $\cos\theta$:
$$\cos\theta = \frac{1}{\sqrt{5}}$$
Finally, we can use the definitions of trigonometric functions to find $\sin\theta$:
$$\sin\theta = \tan\theta\cos\theta = 2\cdot\frac{1}{\sqrt{5}} = \frac{2\sqrt{5}}{5}$$
Therefore, the answer is (B) $5\sqrt{2}/2$.
Given that tan π = 2, ππππ sin π
A. β52β2
B. 5β
2
2
C. β25β5
D. 2β
5
5
3 answers
tanΞΈ = 2
cotΞΈ = 1/2
csc^2ΞΈ = 1 + cot^2ΞΈ = 1 + 1/4 = 5/4
so sinΞΈ = β5/2
cotΞΈ = 1/2
csc^2ΞΈ = 1 + cot^2ΞΈ = 1 + 1/4 = 5/4
so sinΞΈ = β5/2
This is correct! Another way to find $\sin\theta$ is to use the fact that $\cot\theta = 1/2$ is the same as $\tan(\pi/2 - \theta) = 1/2$, which means that $\pi/2 - \theta$ is a reference angle with a ratio of opposite/hypotenuse = 1/2. Since $\pi/2 - \theta$ is acute, we know it must be in the first quadrant, so we can use the Pythagorean Theorem to find the adjacent side:
$$(\text{adjacent})^2 + (\text{opposite})^2 = (\text{hypotenuse})^2$$
$$(\text{adjacent})^2 + (1/2)^2 = 1^2$$
$$(\text{adjacent})^2 = 3/4$$
$$\text{adjacent} = \sqrt{3}/2$$
Therefore, $\sin(\pi/2 - \theta) = \sqrt{1 - (\sqrt{3}/2)^2} = \sqrt{1/4} = 1/2$. But $\sin(\pi/2 - \theta) = \cos\theta$, so we have $\cos\theta = 1/2$. Finally, we can use $\tan^2\theta + 1 = \sec^2\theta$ to find $\sec\theta$ as in the previous solution:
$$\tan^2\theta + 1 = \sec^2\theta$$
$$2^2 + 1 = \sec^2\theta$$
$$\sec\theta = \sqrt{5}$$
Since $\cos\theta = 1/2 > 0$, we know that $\theta$ is either in the first or fourth quadrant. But $\tan\theta = 2 > 0$, which means $\theta$ is in the first quadrant. Therefore, we have $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \sqrt{1 - (\cos\theta)^2} = \sqrt{1 - 1/4} = \boxed{\frac{\sqrt{5}}{2}}$.
$$(\text{adjacent})^2 + (\text{opposite})^2 = (\text{hypotenuse})^2$$
$$(\text{adjacent})^2 + (1/2)^2 = 1^2$$
$$(\text{adjacent})^2 = 3/4$$
$$\text{adjacent} = \sqrt{3}/2$$
Therefore, $\sin(\pi/2 - \theta) = \sqrt{1 - (\sqrt{3}/2)^2} = \sqrt{1/4} = 1/2$. But $\sin(\pi/2 - \theta) = \cos\theta$, so we have $\cos\theta = 1/2$. Finally, we can use $\tan^2\theta + 1 = \sec^2\theta$ to find $\sec\theta$ as in the previous solution:
$$\tan^2\theta + 1 = \sec^2\theta$$
$$2^2 + 1 = \sec^2\theta$$
$$\sec\theta = \sqrt{5}$$
Since $\cos\theta = 1/2 > 0$, we know that $\theta$ is either in the first or fourth quadrant. But $\tan\theta = 2 > 0$, which means $\theta$ is in the first quadrant. Therefore, we have $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \sqrt{1 - (\cos\theta)^2} = \sqrt{1 - 1/4} = \boxed{\frac{\sqrt{5}}{2}}$.