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point (-6,-8) lies on the terminal terminal arm of an angle `theta` in standard...
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You need to evaluate sin theta such that:
`sin theta = y/R`
You may evaluate ` R` using the pythagorean theorem such that:
`R = sqrt(x^2 + y^2) => R = sqrt((-6)^2 + (-8)^2)`
`R = sqrt(36 + 64) => R = 10`
Substituting -8 for y and 10 for R yields:
`sin theta = x/R => sin theta = -8/10 => sin theta = -4/5 => theta = sin^(-1)(-4/5)`
Since sin theta and cos theta are both negative yields that `theta` is an angle in quadrant 3.
`theta = 180^o + 53^o => theta = 233^o`
`cos theta = -6/10 => cos theta = -3/5 => theta = cos^(-1)(-3/5)`
`tan theta = (sin theta)/(cos theta) => tan theta = (-4/5)/(-3/5)`
Reducing duplicate factors yields:
`tan theta = 4/3`
You need to evaluate the sum `sin theta + tan theta` such that:
`sin theta + tan theta = -4/5 + 4/3 => sin theta + tan theta = (-12 + 20)/15 => sin theta + tan theta = 8/15`
Hence, evaluating the sum sin theta + tan theta, under he given conditions, yields `sin theta + tan theta = 8/15` .
Posted by sciencesolve on January 26, 2013 at 7:14 AM (Answer #1)
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