Question

Suppose θ is an angle in the standard position whose terminal side is in Quadrant IV and cotθ = -6/7 . Find the exact values of the five remaining trigonometric functions of θ. Find the exact values of the five remaining trigonometric functions of θ.

Answer 1

let's recall that on the IV Quadrant the sine/y is negative and the cosine/x is positive, whilst the hypotenuse is never negative since it's just a distance unit.

`\bf \stackrel{\textit{on the IV Quadrant}}{cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=√(a^2+b^2) \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=√(6^2+(-7)^2)\implies c=√(36+49)\implies c=√(85) \\\\[-0.35em] ~\dotfill`

`\bf tan(\theta )=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}}\qquad \qquad sec(\theta )=\cfrac{\stackrel{hypotenuse}{√(85)}}{\stackrel{adjacent}{6}}\qquad \qquad csc(\theta )=\cfrac{\stackrel{hypotenuse}{√(85)}}{\stackrel{opposite}{-7}}`

`\bf sin(\theta )=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{√(85)}}\implies \stackrel{\textit{and rationalizing the denominator}}{\cfrac{-7}{√(85)}\cdot \cfrac{√(85)}{√(85)}\implies -\cfrac{7√(85)}{85}} \\\\\\ cos(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{√(85)}}\implies \stackrel{\textit{and rationalizing the denominator}}{\cfrac{6}{√(85)}\cdot \cfrac{√(85)}{√(85)}\implies \cfrac{6√(85)}{85}}`

Answer 2

These are the five remaining trigonometric functions:

• tanθ = - 7/6
• secθ = (√85) / 6
• cosθ = 6(√85) / 85
• sinθ = - 7(√85) / 85
• cscθ = - (√85)/7

Step-by-step explanation:

Quadrant IV corresponds to angle interval 270° < θ < 360.

In this quadrant the signs of the six trigonometric functions are:

• sine and cosecant: negative

• cosine and secant: positive

• tangent and cotangent: negative

The expected values of the five remaining trigonometric functions of θ are:

1) Tangent:

• tan θ = 1 / cot (θ) = 1 / [ -6/7] = - 7/6

2) Secant

• sec²θ = 1 + tan²θ = 1 + (-7/6)² = 1 + 49/36 = 85/36

sec θ = ± (√85)/ 6

Choose positive, because secant is positive in Quadrant IV.

sec θ = (√85) / 6

3) Cosine

• cosθ = 1 / secθ = 6 / (√85) = 6 (√85) / 85

4) Sine

• sin²θ + cos²θ = 1 ⇒ sin²θ = 1 - cos²θ = 1 - [6(√85) / 85] ² =

sin²θ = 1 - 36×85/(85)² = 1- 36/85 = 49/85

sinθ = ± 7 / (√85) = ± 7(√85)/85

Choose negative sign, because it is Quadrant IV.

sinθ = - 7 (√85) / 85

5) Cosecant

• cscθ = 1 / sinθ = - 85 / (7√85) = - (√85) / 7