Mathematics
wwillyshearer
15

Find exact values for sin θ, cos θ and tan θ if sec θ = 6/5 and tan θ < 0 Please show steps as I do not understand this.

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(1) Answers
tammimbari

[latex]\bf sin(\theta)=\cfrac{opposite}{hypotenuse} \qquad \qquad % cosine cos(\theta)=\cfrac{adjacent}{hypotenuse} \\ \quad \\ % tangent tan(\theta)=\cfrac{opposite}{adjacent} \qquad \qquad % secant sec(\theta)=\cfrac{hypotenuse}{adjacent}\\\\ -----------------------------\\\\ sec(\theta)=\cfrac{6}{5}\cfrac{\leftarrow hypotenuse}{\leftarrow adjacent} \\\\\\ [/latex] [latex]\bf \textit{so.. using the pythagorean theorem, we get} \\\\\\ 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} \\\\\\ \textit{that simply means that }\pm \sqrt{6^2-5^2}=b\implies \pm \sqrt{11}=b[/latex] but.. which is it, the positive or negative version of "b"?  well.... we know tangent is < 0, that's another way of saying, tangent is negative now.. .tangent is opposite over adjacent... for tangent to be negative, the sign of those two must differ so.. where does that happens? well, it happens on the 2nd and 4th quadrants, so..... which quadrant then? well, we know the hypotenuse is always positive, is just the radius anyway, and in this case is 6 but the adjacent is positive 5, that means the adjacent is positive, thus the opposite must be negative, and that happens on the 4th quadrant so that means [latex]\bf -\sqrt{11}=b[/latex] so... now, you have all three sides, the hypotenuse, the adjacent, and the opposite, so, just fill those in, in the ratios for cosine, sine and tangent

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