Question 211542

*[Tex \LARGE \sin^2(\theta)+\cos^2(\theta)=1] ... Start with the first Pythagorean identity.



*[Tex \LARGE \left(\frac{1}{4}\right)^2+\cos^2(\theta)=1] ... Plug in *[Tex \LARGE \sin(\theta)=\frac{1}{4}]



*[Tex \LARGE \frac{1}{16}+\cos^2(\theta)=1] ... Square {{{1/4}}} to get {{{1/16}}}



*[Tex \LARGE \cos^2(\theta)=1-\frac{1}{16}] ... Subtract {{{1/16}}} from both sides.



*[Tex \LARGE \cos^2(\theta)=\frac{16}{16}-\frac{1}{16}] ... Multiply 1 by {{{16/16}}}



*[Tex \LARGE \cos^2(\theta)=\frac{16-1}{16}] ... Combine the fractions.



*[Tex \LARGE \cos^2(\theta)=\frac{15}{16}] ... Subtract



*[Tex \LARGE \cos(\theta)=\sqrt{\frac{15}{16}}] ... Take the square root of both sides. Note: because *[Tex \Large \theta] is in the first quadrant, this means that *[Tex \Large \cos(\theta)>0] (ie cosine is positive). So we don't need to worry about the negative square root.



*[Tex \LARGE \cos(\theta)=\frac{\sqrt{15}}{\sqrt{16}}] ... Break up the square root.



*[Tex \LARGE \cos(\theta)=\frac{\sqrt{15}}{4}] ... Take the square root of 16 to get 4.



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*[Tex \LARGE \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}] ... Move onto the tangent identity.



*[Tex \LARGE \tan(\theta)=\frac{\frac{1}{4}}{\frac{\sqrt{15}}{4}}] ... Plug in *[Tex \LARGE \sin(\theta)=\frac{1}{4}] and *[Tex \LARGE \cos(\theta)=\frac{\sqrt{15}}{4}]



*[Tex \LARGE \tan(\theta)=\frac{1}{4}\cdot\frac{4}{\sqrt{15}}] ... Multiply the first fraction by the reciprocal of the second fraction.



*[Tex \LARGE \tan(\theta)=\frac{1}{\sqrt{15}}] ... Cancel out the common terms and simplify.



*[Tex \LARGE \tan(\theta)=\frac{1}{\sqrt{15}}\cdot\frac{\sqrt{15}}{\sqrt{15}}] ... Multiply the fraction by {{{sqrt(15)/sqrt(15)}}} (this will rationalize the denominator).



*[Tex \LARGE \tan(\theta)=\frac{1\cdot\sqrt{15}}{\sqrt{15}\cdot\sqrt{15}}] ... Combine the fractions.



*[Tex \LARGE \tan(\theta)=\frac{\sqrt{15}}{15}] ... Multiply



So if *[Tex \LARGE \sin(\theta)=\frac{1}{4}], then *[Tex \LARGE \tan(\theta)=\frac{\sqrt{15}}{15}]