Question

sinθ = 4/5 cscθ = 5/4
cosθ = 3/5 secθ = 5/3
tanθ = 4/3 cotθ = 3/4

1. draw a diagram and find the exact,

simplified values given that tanβ =
3 and the terminal side of β is in
quadrant i.
sinβ = 3√10/10 cotβ = 1/3

Explanation:

Step1: Recall the definition of tangent

Since \(\tan\beta=\frac{y}{x} = 3=\frac{3}{1}\) (in the first - quadrant, \(x>0,y>0\)), we can assume \(x = 1\) and \(y=3\).

Step2: Use the Pythagorean theorem to find \(r\)

By the Pythagorean theorem \(r=\sqrt{x^{2}+y^{2}}\), substituting \(x = 1\) and \(y = 3\), we get \(r=\sqrt{1^{2}+3^{2}}=\sqrt{1 + 9}=\sqrt{10}\).

Step3: Calculate \(\sin\beta\)

Using the formula \(\sin\beta=\frac{y}{r}\), substituting \(y = 3\) and \(r=\sqrt{10}\), we have \(\sin\beta=\frac{3}{\sqrt{10}}=\frac{3\sqrt{10}}{10}\) (rationalizing the denominator).

Step4: Calculate \(\cos\beta\)

Using the formula \(\cos\beta=\frac{x}{r}\), substituting \(x = 1\) and \(r=\sqrt{10}\), we get \(\cos\beta=\frac{1}{\sqrt{10}}=\frac{\sqrt{10}}{10}\).

Step5: Calculate \(\csc\beta\)

Since \(\csc\beta=\frac{1}{\sin\beta}\), substituting \(\sin\beta=\frac{3\sqrt{10}}{10}\), we have \(\csc\beta=\frac{\sqrt{10}}{3}\).

Step6: Calculate \(\sec\beta\)

Since \(\sec\beta=\frac{1}{\cos\beta}\), substituting \(\cos\beta=\frac{\sqrt{10}}{10}\), we get \(\sec\beta=\sqrt{10}\).

Step7: Calculate \(\cot\beta\)

Since \(\cot\beta=\frac{1}{\tan\beta}\), substituting \(\tan\beta = 3\), we have \(\cot\beta=\frac{1}{3}\).

Answer:

\(\sin\beta=\frac{3\sqrt{10}}{10},\cos\beta=\frac{\sqrt{10}}{10},\tan\beta = 3,\csc\beta=\frac{\sqrt{10}}{3},\sec\beta=\sqrt{10},\cot\beta=\frac{1}{3}\)