If sin(theta)=[sqrt(70)]/7 and theta is in Quadrant two, find the exact numerical value of tan theta without using a calculator.

I got tan(theta)=[sqrt(294)]/42

Is that right?

if sin(theta)=-a where 0<a<1, and theta is in quadrant 3, find the exact algebraic expressionm for cos(theta)

You said

sin(theta)=[sqrt(70)]/7
that would make
sin(theta) > 1 which is not possible

check your typing

To find the exact numerical value of tan(theta) without using a calculator, we can use the relationship between sine and tangent.

We know that sin(theta) = sqrt(70)/7. Since the given value of sin(theta) is positive, we can determine the cosine of theta using the Pythagorean identity: sin^2(theta) + cos^2(theta) = 1.

Let's solve for cos(theta):
sin^2(theta) + cos^2(theta) = 1
(sqrt(70)/7)^2 + cos^2(theta) = 1
70/49 + cos^2(theta) = 1
cos^2(theta) = 1 - 70/49
cos^2(theta) = 49/49 - 70/49
cos^2(theta) = -21/49
cos(theta) = sqrt(-21)/7

Since theta is in Quadrant two, both sine and cosine are positive. Therefore, we can keep the positive square root in the expression for cos(theta). Thus, the exact numerical value of cos(theta) is sqrt(-21)/7.

Now, we can find the tangent of theta using the identity tan(theta) = sin(theta) / cos(theta):
tan(theta) = (sqrt(70)/7) / (sqrt(-21)/7)
tan(theta) = sqrt(70)/sqrt(-21)
To simplify this expression, multiply the numerator and denominator by sqrt(21):
tan(theta) = (sqrt(70) * sqrt(21)) / (sqrt(-21) * sqrt(21))
tan(theta) = sqrt(70 * 21) / sqrt(-21 * 21)
tan(theta) = sqrt(1470) / sqrt(441)
tan(theta) = sqrt(1470) / 21

Therefore, the exact numerical value of tan(theta) without using a calculator is sqrt(1470) / 21.

Regarding the second question:
If sin(theta) = -a, where 0 < a < 1, and theta is in Quadrant 3, we can use the definition of cosine to find its algebraic expression.

In Quadrant 3, both sine and cosine are negative. So, we can say that cos(theta) = -sqrt(1 - sin^2(theta)). Substituting sin(theta) = -a into this expression, we get:
cos(theta) = -sqrt(1 - (-a)^2)
cos(theta) = -sqrt(1 - a^2)
Hence, the exact algebraic expression for cos(theta) is -sqrt(1 - a^2).