solve sin(theta)=-1 for all real values of (theta)
I don't understand but by following the example in my book;
sec(theta)=-1
from the pattern of the secant function, sec(theta)=-1 if theta=pi*n, where n is an odd integer.
There is no example of sin.
Havent you ever seen a sine or cosine curve? Surely that is in your book. If not, toss the book quickly.
http://en.wikipedia.org/wiki/Sine_wave
thank you for the links and everything but I need someone outside the computer world to explain
my previous post still stands but is the answer 5pi/2?
To solve sin(theta) = -1 for all real values of theta, we can use the unit circle or trigonometric identities.
Using the unit circle:
1. Recall that sin(theta) represents the y-coordinate of a point on the unit circle at angle theta.
2. The unit circle shows that sin(theta) is equal to -1 when theta is equal to 3π/2 plus any multiple of 2π. This means that theta can be written as: theta = (3π/2) + 2πn, where n is an integer.
Using trigonometric identities:
1. Recall the identity: sin(theta) = -1.
2. Rewrite the identity as: sin(theta) + 1 = 0.
3. Factor out sin(theta) + 1 by using the difference of squares: (sin(theta) + 1)(sin(theta) - 1) = 0.
4. Set each factor equal to zero: sin(theta) + 1 = 0 and sin(theta) - 1 = 0.
5. Solve each equation separately:
- For sin(theta) + 1 = 0, subtract 1 from both sides to get sin(theta) = -1.
- For sin(theta) - 1 = 0, add 1 to both sides to get sin(theta) = 1.
6. Focus on the sin(theta) = -1 equation. By referring to the unit circle or the pattern of the sine function, we can determine that theta is equal to 3π/2 plus any multiple of 2π.
- Therefore, theta = (3π/2) + 2πn, where n is an integer.
In conclusion, the solution to sin(theta) = -1 for all real values of theta is theta = (3π/2) + 2πn, where n is an integer.