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lapo4ka [179]
3 years ago
5

Suppose θ is an angle in the fourth quadrant with cosθ=x/4. Find expressions for the other five trig functions in terms of x?

Mathematics
1 answer:
IrinaVladis [17]3 years ago
6 0
\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\quad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\\\\\\
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\qquad 
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\quad 
% secant
sec(\theta)=\cfrac{hypotenuse}{adjacent}\\\\
-------------------------------

now, we know the angle is in the IV quadrant, that means, the cosine is positive and the sine is negative, or "x" is positive whilst "y" is negative.

\bf cos(\theta )=\cfrac{\stackrel{adjacent}{x}}{\stackrel{hypotenuse}{4}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{4^2-x^2}=b\implies \stackrel{IV~quadrant}{-\sqrt{16-x^2}=b}\\\\
-------------------------------

\bf sin(\theta)=\cfrac{-\sqrt{16-x^2}}{4}
\qquad
cos(\theta)=\cfrac{x}{4}
\qquad 
% tangent
tan(\theta)=\cfrac{-\sqrt{16-x^2}}{x}
\\\\\\
% cotangent
cot(\theta)=\cfrac{x}{-\sqrt{16-x^2}}
\qquad 
% cosecant
csc(\theta)=\cfrac{4}{-\sqrt{16-x^2}}
\qquad 
% secant
sec(\theta)=\cfrac{4}{x}

now, for the cotangent and the cosecant, let's rationalize the denominator,

\bf cot(\theta)=\cfrac{x}{-\sqrt{16-x^2}}\cdot \cfrac{\sqrt{16-x^2}}{\sqrt{16-x^2}}\implies cot(\theta)=-\cfrac{x\sqrt{16-x^2}}{16-x^2}
\\\\\\
csc(\theta)=\cfrac{4}{-\sqrt{16-x^2}}\cdot \cfrac{\sqrt{16-x^2}}{\sqrt{16-x^2}}\implies csc(\theta)=-\cfrac{4\sqrt{16-x^2}}{16-x^2}
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The question is incomplete. Here is the complete question

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I'm guessing the series is supposed to be

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