If secx = 8 and -pi/2 < x < 0, find the exact value of sin2x
Use the identity sin 2x = 2(sinx)(cosx)
if secx = 8, then cosx = 1/8 where x is in the fourth quadrant.
consider a right angled triangle with x=1, r=8, then y=??
by Pythagoras 1^2 + y^2 = 8^2
y = -√63 (it has to be negative in IV)
so sinx = -√63/8
so we have sin 2x = 2(-√63/8)(1/8)
=-√63/32
Why was the math book sad?
Because it had too many problems!
But don't worry, I can help you solve this one. The exact value of sin 2x is -√63/32.
To find the exact value of sin2x, we'll use the identity sin2x = 2(sinx)(cosx).
Given that secx = 8 and -π/2 < x < 0, we can determine that cosx = 1/8.
Since x is in the fourth quadrant, consider a right-angled triangle with x=1, r=8. Using the Pythagorean theorem, we have 1^2 + y^2 = 8^2. Solving for y, we get y = -√63 (since it has to be negative in the fourth quadrant).
Therefore, sinx = -√63/8.
Now, substituting sinx and cosx into the formula, sin2x = 2(-√63/8)(1/8), which simplifies to -√63/32.
Thus, the exact value of sin2x is -√63/32.
To find the exact value of sin 2x, we can use the identity sin 2x = 2(sinx)(cosx).
Given that secx = 8 and -pi/2 < x < 0, we can first find cosx. Since secx is the reciprocal of cosx, we have secx = 1/cosx = 8. Solving for cosx, we get cosx = 1/8.
Since our angle x is in the fourth quadrant (where cosine is positive), we can consider a right-angled triangle with the x-coordinate as 1, the radius as 8, and solve for the y-coordinate. Using the Pythagorean theorem, we have 1^2 + y^2 = 8^2. Solving for y, we get y = -√63. It is negative in the fourth quadrant.
Thus, we have sinx = y/r = -√63/8.
Now, substituting these values into the sin 2x formula, we get:
sin 2x = 2(-√63/8)(1/8) = -√63/32.
Therefore, the exact value of sin 2x is -√63/32.