Respuesta :
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.
[tex]\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=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf tan(\theta )=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}}\qquad \qquad sec(\theta )=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{adjacent}{6}}\qquad \qquad csc(\theta )=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{opposite}{-7}}[/tex]
[tex]\bf sin(\theta )=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{\cfrac{-7}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies -\cfrac{7\sqrt{85}}{85}} \\\\\\ cos(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{\cfrac{6}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies \cfrac{6\sqrt{85}}{85}}[/tex]
Answer:
These are the five remaining trigonometric functions:
- tanθ = - 7/6
- secθ = (√85) / 6
- cosθ = 6(√85) / 85
- sinθ = - 7(√85) / 85
- cscθ = - (√85)/7
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
