Breabel
contestada

Suppose sin theta=x/9 and the angle theta is in the 1st quadrant. Write algebraic expressions for cos(theta) and tan(theta) in terms of x.

cos(theta)=

tan(theta)=

Respuesta :

arthurpdc
We'll consider that x > 0. By the Pythagorean trigonometric identity, we have:

[tex]\sin^2\theta+\cos^2\theta=1\\\\ \left(\dfrac{x}{9}\right)^2+\cos^2\theta=1\\\\ \dfrac{x^2}{81}+\cos^2\theta=1\\\\ \cos^2\theta=1-\dfrac{x^2}{81}\\\\ \cos^2\theta=\dfrac{81-x^2}{81}\\\\ \cos\theta=\pm\sqrt{\dfrac{81-x^2}{81}}\\\\\cos\theta=\pm\dfrac{\sqrt{81-x^2}}{9}[/tex]

Since θ is in the 1st quadrant, [tex]\cos\theta\ \textgreater \ 0[/tex]. So:

[tex]\boxed{\cos\theta=\dfrac{\sqrt{81-x^2}}{9}}[/tex]

Now, we'll find the tangent. Using the formula below:

[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}\\\\ \tan\theta=\dfrac{\dfrac{x}{9}}{~~\dfrac{\sqrt{81-x^2}}{9}~~}=\dfrac{x}{9}\times\dfrac{9}{\sqrt{81-x^2}}\\\\ \boxed{\tan\theta=\dfrac{x}{\sqrt{81-x^2}}}[/tex]

Otras preguntas