given that sec theta = radical13/whole number 3, and csctheta<0 find sin theta using the 2different methods;
1)using the "x-y-r" definition
2) using identities only
Timmy, Ryland, Cobra, Jaime -- you must be having an identity crisis tonight.
Please use the same name for all of your posts.
given sec Ø = √13/3 ---> cosØ = 3/√13
also cscØ < 0 ---> sinØ < 0
Ø must be in quadr IV
so x = 3 , r = √13
x^2 + y^2 = r^2
9 + y^2 = 13
y^2 = 4
y = ± 2 , but in IV y = -2
sin Ø = -2/√13
2) you should know that
sin^2Ø + cos^2Ø = 1
sin^2Ø + 9/13 = 1
sin^2Ø = 4/13
sinØ = 2/√13 , but we know that Ø is in IV, where the sine is negative, so
sinØ = -2/√13
To find sin theta using the given information, let's solve the problem step by step using two different methods.
Method 1: Using the "x-y-r" definition
Step 1: Start with the given information:
sec theta = √13/3
Step 2: Rewrite sec theta in terms of sin theta:
sec theta = 1/cos theta
1/cos theta = √13/3
Step 3: Multiply both sides of the equation by cos theta:
1 = (√13/3) * cos theta
Step 4: Solve for cos theta:
cos theta = 3/√13
Step 5: Use the Pythagorean identity to find sin theta:
sin theta = ±√(1 - cos^2 theta)
sin theta = ±√(1 - (3/√13)^2)
sin theta = ±√(1 - 9/13)
sin theta = ±√(4/13)
sin theta = ±2/√13
Since csctheta < 0, sin theta must be negative, so:
sin theta = -2/√13
Method 2: Using identities only
Step 1: Start with the given information:
sec theta = √13/3
Step 2: Use the identity sec theta = 1/cos theta:
1/cos theta = √13/3
Step 3: Cross-multiply the equation:
3 = √13 * cos theta
Step 4: Square both sides of the equation:
9 = 13 * cos^2 theta
Step 5: Solve for cos^2 theta:
cos^2 theta = 9/13
Step 6: Use the Pythagorean identity to find sin theta:
sin^2 theta = 1 - cos^2 theta
sin^2 theta = 1 - 9/13
sin^2 theta = 4/13
Step 7: Take the square root of both sides to find sin theta:
sin theta = ±√(4/13)
sin theta = ±2/√13
Since csctheta < 0, sin theta must be negative, so:
sin theta = -2/√13
In both methods, we find that sin theta is equal to -2/√13.