Question

in which quadrant are all trigonometric functions positive? use 1, 2, 3 or 4 for your answer. (1 point) all trigonometric functions are positive in quadrant.

Explanation:

Step1: Recall trigonometric sign rules

In the coordinate plane, for an angle in standard position with a point \((x,y)\) on its terminal side and \(r = \sqrt{x^{2}+y^{2}}\):

• Sine function: \(\sin\theta=\frac{y}{r}\)
• Cosine function: \(\cos\theta=\frac{x}{r}\)
• Tangent function: \(\tan\theta=\frac{y}{x}\) (for \(x

eq0\))

• Cosecant function: \(\csc\theta=\frac{r}{y}\) (for \(y

eq0\))

• Secant function: \(\sec\theta=\frac{r}{x}\) (for \(x

eq0\))

• Cotangent function: \(\cot\theta=\frac{x}{y}\) (for \(y

eq0\))

Step2: Analyze quadrants

• Quadrant 1: \(x>0\), \(y>0\), \(r>0\) (since \(r = \sqrt{x^{2}+y^{2}}\) and \(x,y\) positive). So \(\sin\theta=\frac{y}{r}>0\), \(\cos\theta=\frac{x}{r}>0\), \(\tan\theta=\frac{y}{x}>0\), \(\csc\theta=\frac{r}{y}>0\), \(\sec\theta=\frac{r}{x}>0\), \(\cot\theta=\frac{x}{y}>0\) (all positive).
• Quadrant 2: \(x<0\), \(y>0\), \(r>0\). \(\sin\theta=\frac{y}{r}>0\), \(\cos\theta=\frac{x}{r}<0\), so not all positive.
• Quadrant 3: \(x<0\), \(y<0\), \(r>0\). \(\tan\theta=\frac{y}{x}>0\) (negative/negative), but \(\sin\theta=\frac{y}{r}<0\), \(\cos\theta=\frac{x}{r}<0\), so not all positive.
• Quadrant 4: \(x>0\), \(y<0\), \(r>0\). \(\cos\theta=\frac{x}{r}>0\), but \(\sin\theta=\frac{y}{r}<0\), so not all positive.

Answer:

1