cos(-7pi/2)=cos(pi+pi/2)
Yes, this statement is true.
We know that cos(-θ) = cos(θ). Therefore, cos(-7π/2) = cos(7π/2).
Also, cos(π + θ) = -cos(θ). Therefore, cos(π + π/2) = -cos(π/2).
And since cos(π/2) = 0, we have -cos(π/2) = -0 = 0.
So, cos(-7π/2) = cos(7π/2) = cos(π + π/2) = 0.
To simplify the expression cos(-7π/2), we need to use the properties of cosine function.
The cosine function has a period of 2π, meaning that cos(x) = cos(x + 2π) for any value of x.
So, we can add or subtract multiples of 2π to simplify the expression in a way that makes it easier to calculate.
In this case, we can add 2π to -7π/2 to get:
-7π/2 + 2π = -7π/2 + 4π/2 = -3π/2
Now, we can simplify cos(-3π/2) by recognizing that cosine is negative in the second and third quadrants.
In the unit circle, the angle -3π/2 corresponds to a point on the negative y-axis.
Since the cosine function represents the x-coordinate of that point, we know that cos(-3π/2) = 0.
Therefore, cos(-7π/2) is equal to 0.
Now let's compare it with cos(pi + pi/2).
To simplify cos(pi + pi/2), we add the angles together:
pi + pi/2 = (2pi/2) + (pi/2) = (3pi/2)
Similar to the previous calculation, cos(3π/2) is equal to 0.
So, we can conclude that cos(-7π/2) is equal to cos(pi + pi/2) and both values are 0.