Find the Other Trig Values in Quadrant II sin(x)=( square root of 5)/7

Problem

sin(x)=√(,5)/7, Quadrant II

Solution

1. Identify the given information and the quadrant. In Quadrant II, the sine value is positive, while cosine, tangent, cotangent, and secant are negative.

2. Use the Pythagorean identity to find cos(x)

sin2(x)+cos2(x)=1

3. Substitute the known value of sin(x) into the identity.

(√(,5)/7)2+cos2(x)=1

4. Solve for cos2(x)

5/49+cos2(x)=1

cos2(x)=1−5/49

cos2(x)=44/49

5. Take the square root and apply the sign for Quadrant II. Since cosine is negative in Quadrant II:

cos(x)=−√(,44/49)

cos(x)=−(2√(,11))/7

6. Find the tangent using the ratio of sine to cosine.

tan(x)=sin(x)/cos(x)

tan(x)=√(,5)/7/(−(2√(,11))/7)

tan(x)=−√(,5)/(2√(,11))

tan(x)=−√(,55)/22

7. Find the cosecant by taking the reciprocal of sine.

csc(x)=1/sin(x)

csc(x)=7/√(,5)

csc(x)=(7√(,5))/5

8. Find the secant by taking the reciprocal of cosine.

sec(x)=1/cos(x)

sec(x)=−7/(2√(,11))

sec(x)=−(7√(,11))/22

9. Find the cotangent by taking the reciprocal of tangent.

cot(x)=1/tan(x)

cot(x)=−(2√(,11))/√(,5)

cot(x)=−(2√(,55))/5

Final Answer

cos(x)=−(2√(,11))/7,tan(x)=−√(,55)/22,csc(x)=(7√(,5))/5,sec(x)=−(7√(,11))/22,cot(x)=−(2√(,55))/5

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