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  • 5 years ago

I'm given Cos(theta) = -2/3 and Tan(theta)=1/2 *sqrt(5) and I'm asked to find sin, sec, csc, cot,

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  1. anonymous
    • 5 years ago
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    Think of a right triangle, draw it. You have the long side (Hypotenuse) and you have the short sides (the legs). Put a Theta symbol in one of the two non-90degree angles. The short side across from your theta symbol we will call O (O for opposite). The short side (not the hypotenuse) that is next to your theta symbol we will call A (A for adjacent). The hypotenuse we will denote as H. The following trig functions can be defined as follows: sin(theta)=O/H csc(theta)=1/sin(theta) cos(theta)=A/H sec(theta)=1/cos(theta) tan(theta)=O/A cot(theta)=1/tan(theta) If cos(theta) = -2/3, and cos(theta)=A/H, then you know that A=-2 and H=3. Use the Pythagorean theorem (a^2+b^2=c^2), in this case, (A^2+O^2=H^2) to find the last side. Once you know the last side, you can finish defining the last remaining trig functions.

  2. anonymous
    • 5 years ago
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    That's very helpful. However the way the question is given it doesn't allow for such simple calculations. It asks for the quadrant which I found to be 3 then using your method I've been marked incorrect. Its the tan(theta)= 1/2 * sqrt(5) that is throwing me off.

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