How Does Quadrant Affect Values in Trigonometric Functions?

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To find the cosine of an angle where tan(b) = -√3 and b lies in quadrant III, the sine value is calculated as sin(b) = -√3/2. The cosine can be derived using the relationship cos(b) = x/r, where x is the adjacent side and r is the hypotenuse. However, there is confusion regarding the quadrant placement, as tan(b) being negative suggests b could be in quadrant II or IV, not III. The correct cosine value must be determined based on accurate quadrant identification and the corresponding sine value. Understanding the relationship between sine, cosine, and tangent is crucial for solving trigonometric functions accurately.
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How would I find cos for tan when..

Tan b = (-√3)

Theta lies in quadrant iii

Sin b would equal (-√3) / 2

So Cos b would equal -1 / 2?
 
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y: -√3
x: 1
r: √2 > from x2 + y2

cos = x/r

Answer: ?
 
CrossFit415 said:
How would I find cos for tan when..

Tan b = (-√3)

Theta lies in quadrant iii
What does theta have to do with anything? Assuming that you meant b instead of theta, if tan(b) = -√3, then b is in quadrant II or quadrant IV.
CrossFit415 said:
Sin b would equal (-√3) / 2

So Cos b would equal -1 / 2?
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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