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)
If sin(theta)=[sqrt(7)]/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 posted the same question earlier today, and I told you .....
http://www.jiskha.com/display.cgi?id=1231449736
yes, but i changed my mistake from
sin(theta)=[sqrt(70)]/7 to sin(theta)=[sqrt(7)]/7.
so it's sin(theta)=.378
ok then, that's better
recall that sine(angle) = opposite/hypotenus
so we need a right-angled triangle in the II quadrant with a height of √7 and a hypotenuse of 7
let the base be x
x^2 + (√7)^2 = 7^2
x^2 = 42
x = ±√42, but we are in the second quadrant so x = -√42
then for yours
tan(theta) = √7/-√42
= -1/√6
notice all steps were done without a calculator.
that does not mean we couldn't use a calculator to check our answer
enter the following
√7/7 =
2nd function sin
-180=
± key (to make our answer positive in the second quadrant)(on some calculators you might have to multiply by -1 to get it to a positive, do whatever your calc needs done)
tan =
store or write down that number
now do
-1/√6 =
compare the two results, they are the same.
To find the exact numerical value of tan(theta) without using a calculator, you can use the relationship between sin(theta) and tan(theta) in the given quadrant.
In Quadrant 2, both sin(theta) and tan(theta) are positive.
Given sin(theta) = (√70)/7, we can analyze the triangle in Quadrant 2 to find the other trigonometric values.
From sin(theta) = (√70)/7, we know that the opposite side is (√70) and the hypotenuse is 7. Using the Pythagorean theorem (a^2 + b^2 = c^2), we can find the adjacent side:
a^2 + (√70)^2 = 7^2
a^2 + 70 = 49
a^2 = 49 - 70
a^2 = -21
Since theta is in Quadrant 2, the adjacent side is negative. Therefore, a = -√21.
Now, we can use the formula for tan(theta): tan(theta) = opposite/adjacent.
tan(theta) = (√70)/(-√21)
To rationalize the denominator, we can multiply both the numerator and denominator by √21:
tan(theta) = (√70 * √21) / (-√21 * √21)
tan(theta) = (√(70*21)) / (-√(21*21))
tan(theta) = (√(1470)) / (-21)
tan(theta) = (√(1470))/(-21)
Therefore, the exact numerical value of tan(theta) without using a calculator is (√(1470))/(-21).
For the second part of your question, when sin(theta) = -a where 0 < a < 1, and theta is in Quadrant 3, we need to find the expression for cos(theta).
In Quadrant 3, both sin(theta) and cos(theta) are negative.
Given sin(theta) = -a, we can use the Pythagorean identity to find the expression for cos(theta):
cos^2(theta) = 1 - sin^2(theta)
Substituting -a for sin(theta), we get:
cos^2(theta) = 1 - (-a)^2
cos^2(theta) = 1 - a^2
cos(theta) = -√(1 - a^2)
Therefore, the exact algebraic expression for cos(theta) when sin(theta) = -a and theta is in Quadrant 3 is -√(1 - a^2).