Question

1. a right triangle has a hypotenuse of length 3√2 and a leg of length √5. use the pythagorean theorem to find the exact value of the other leg. then let θ be the angle opposite the shortest side, then find the six trigonometric functions of the angle θ. a² + b² = c² (√5)² + b² = (3√2)² 8. use the even - odd and periodic properties to simplify: os(−x)+2cos(x + 4π)

Explanation:

Step1: Find the other leg

Let the known leg $a = \sqrt{5}$ and the hypotenuse $c=3\sqrt{2}$. By the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, so $(\sqrt{5})^{2}+b^{2}=(3\sqrt{2})^{2}$. Then $5 + b^{2}=18$, and $b^{2}=18 - 5=13$, so $b=\sqrt{13}$. Since $\sqrt{5}<\sqrt{13}$, the side of length $\sqrt{5}$ is the shortest side.

Step2: Define the six - trigonometric functions

The six trigonometric functions of an angle $\theta$ in a right - triangle are defined as follows:

• Sine: $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\sqrt{5}}{3\sqrt{2}}=\frac{\sqrt{10}}{6}$
• Cosine: $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt{13}}{3\sqrt{2}}=\frac{\sqrt{26}}{6}$
• Tangent: $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sqrt{5}}{\sqrt{13}}=\frac{\sqrt{65}}{13}$
• Cosecant: $\csc\theta=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{3\sqrt{2}}{\sqrt{5}}=\frac{3\sqrt{10}}{5}$
• Secant: $\sec\theta=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{3\sqrt{2}}{\sqrt{13}}=\frac{3\sqrt{26}}{13}$
• Cotangent: $\cot\theta=\frac{\text{adjacent}}{\text{opposite}}=\frac{\sqrt{13}}{\sqrt{5}}=\frac{\sqrt{65}}{5}$

For the second part:

Step1: Use even - odd and periodic properties

The cosine function is even, so $\cos(-x)=\cos x$. The cosine function has a period of $2\pi$, so $\cos(x + 4\pi)=\cos x$.
Then $\cos(-x)+2\cos(x + 4\pi)=\cos x+2\cos x = 3\cos x$

Answer:

For the right - triangle problem:
$\sin\theta=\frac{\sqrt{10}}{6},\cos\theta=\frac{\sqrt{26}}{6},\tan\theta=\frac{\sqrt{65}}{13},\csc\theta=\frac{3\sqrt{10}}{5},\sec\theta=\frac{3\sqrt{26}}{13},\cot\theta=\frac{\sqrt{65}}{5}$
For the simplification problem: $3\cos x$