If cos(theta)=-1/3 and theta is in the interval [pi, 3pi/2], then find the value of sin(theta).

Question

Please help with this question!

Answer 1

If cos(theta)=-1/3 and theta is in the interval [pi, 3pi/2], then find the value of sin(theta).

 

Use the pythagorean identity!  square of cosine is 1/9, so square of sine must be 8/9, so sine must be

NEGATIVE √8 / 3, because in quadrant iii both sine & cosine are negative.

 

I regard this as a superior method although it's equivalent to the pseudo-geometric way (with negative values for 'opposite' and 'adjacent' sides); both are based on Pythagorean Theorem.

 

Answer: -2√2 ÷ 3 when simplified 

If cos(theta)=-1/3 and theta is in the interval [pi, 3pi/2], then find the value of sin(theta)

Answer 2

Hey Catherine

if theta is int the 3th quadrant then we have both adjacent and opposite negative

now from cos theta =-1/3 we can deduce that adjacent is -1 and hypotenuse is 3

Applying Pythagorean theorem we can get the opposite

so (opposite)^2+(-1)^2=(3)^2

opposite= ±√[3^2-1]=±√8=±2√2

we already said it is negative so -2√2 is good

then, sin (theta) = -2√2/3

Hope that helps!