Question

find sinϴ and cosϴ if tanϴ=1/4 and sinϴ>0

Answer 1

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$\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\ \mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\ \mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}$


Square both sides:

$\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\ \mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\ \mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\ \mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}$

$\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\ \mathsf{17\,sin^2\,\theta=1}\\\\ \mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\ \mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\ \mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}$


Since $\mathsf{sin\,\theta}$ is positive, you can discard the negative sign. So,

$\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}$


Substitute this value back into $\mathsf{(i)}$ to find $\mathsf{cos\,\theta:}$

$\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\ \mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}$


I hope this helps. =)


Tags:   trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus