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Suppose sin(a)=3/4

use the trig identity sin^2(a)+cos^2(a)=1
and the trig identity tan(a)=sin(a)/cos(a)
to find tan (a) in quad II.
Round to the nearest hundredth.

1 Answer

3 votes

well, we know the sine, and we also know that we're on the II Quadrant, let's recall that on the II Quadrant sine is positive whilst cosine is negative.


\bf sin^2(\theta)+cos^2(\theta)=1~\hspace{10em} tan(\theta )=\cfrac{sin(\theta )}{cos(\theta )} \\\\[-0.35em] ~\dotfill\\\\ sin^2(a)+cos^2(a)=1\implies cos^2(a) = 1-sin^2(a) \\\\\\ cos^2(a) = 1-[sin(a)]^2\implies cos^2(a) = 1-\left( \cfrac{3}{4} \right)^2\implies cos^2(a) = 1-\cfrac{3^2}{4^2} \\\\\\ cos^2(a) = 1-\cfrac{9}{16}\implies cos^2(a) = \cfrac{7}{16}\implies cos(a)=\pm\sqrt{\cfrac{7}{16}}


\bf cos(a)=\pm\cfrac{√(7)}{√(16)}\implies cos(a)=\pm\cfrac{√(7)}{4}\implies \stackrel{\textit{on the II Quadrant}}{cos(a)=-\cfrac{√(7)}{4}}\\\\[-0.35em]~\dotfill\\\\tan(a)=\cfrac{sin(a)}{cos(a)}\implies tan(a)=\cfrac{~~(3)/(4)~~}{-(√(7))/(4)}\implies tan(a)=\cfrac{3}{4}\cdot \cfrac{4}{-√(7)}\\\\\\tan(a)=-\cfrac{3}{√(7)}\implies \stackrel{\textit{rounded up}}{tan(a) = -1.13}

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