Use an appropriate identity to find the exact value cos(-7pi/12)
wrong again.
cos(-7π/6) = -√3/2
so, cos(-7π/12) = -√((1 - √3/2)/2) = -(√3-1)/(2√2)
We know that cosine is an even function, which means that cos(x) = cos(-x) for all real numbers x.
Therefore, cos(-7π/12) = cos(7π/12).
To find the exact value of cos(7π/12), we can use the half-angle identity for cosine:
cos(θ/2) = ±√((1 + cos(θ))/2)
Using this identity, we can express cos(7π/12) in terms of cos(π/6) or cos(π/3), which are known values.
Since 7π/12 is closer to π/3 than π/6 on the unit circle, we'll express cos(7π/12) in terms of cos(π/3):
cos(7π/12) = cos((2π/3)/2) = ±√((1 + cos(2π/3))/2)
In the unit circle, cos(2π/3) = -1/2.
Plug in this value to the equation above:
cos(7π/12) = ±√((1 - 1/2)/2) = ±√((1/2)/2) = ±√(1/4) = ±1/2
Therefore, the exact value of cos(-7π/12) is ±1/2.
Apologies for the mistake in my previous response.
To find the exact value of cos(-7π/12), we can use the double-angle identity for cosine:
cos(2θ) = 2cos^2(θ) - 1
Using this identity, we can express cos(-7π/12) in terms of cos(7π/6), which is a known value.
cos(-7π/12) = cos(2(7π/12)) = 2cos^2(7π/12) - 1
Now, cos(7π/6) = -√3/2. We can use this value to calculate cos^2(7π/12).
cos^2(7π/12) = (cos(7π/6))^2 = (-√3/2)^2 = 3/4
Plugging this into the previous equation:
cos(-7π/12) = 2(3/4) - 1 = 3/2 - 1 = 1/2
Therefore, the exact value of cos(-7π/12) is 1/2.