Write y= -2+7csc(3x-6) in terms of secant.
y= -2+7csc(3x-6)
csc(3x-6) = (y+2)/7
or sin(3x-6) = 7/(y+2)
construct a right-angled triangle with angle (3x-6), opposite side 7, and hypotenuse Y+2
then x^2 + 49 = (y+2)^2
x = √((y+2)^2 - 49)
cos(3x-6) = √((y+2)^2 - 49) / (y+2)
sec(3x-6) = (y+2)/ √((y+2)^2 - 49)
if you just want to replace the csc by sec, then since
cscu = secu/√(sec^2u-1)
-2+7csc(3x-6) = -2+7(sec(3x-6)/√(sec^2(3x-6)-1)
To express the given equation y = -2 + 7csc(3x-6) in terms of secant, we need to use trigonometric identities to write csc(3x-6) in terms of sec(3x-6), then substitute it in the equation.
First, let's recall the reciprocal identity for csc(x):
csc(x) = 1/sin(x)
Now, we can write csc(3x-6) in terms of sin(3x-6):
csc(3x-6) = 1/sin(3x-6)
Next, we need to express sin(3x-6) in terms of sec(3x-6). To do this, we will use the Pythagorean Identity:
sin^2(x) + cos^2(x) = 1
Rearranging this equation, we get:
sin^2(x) = 1 - cos^2(x)
Substituting (3x-6) for x, we have:
sin^2(3x-6) = 1 - cos^2(3x-6)
Now, divide both sides by cos^2(3x-6):
sin^2(3x-6)/cos^2(3x-6) = (1 - cos^2(3x-6))/cos^2(3x-6)
Simplifying the left side using the Quotient Identity:
tan^2(3x-6) = (1 - cos^2(3x-6))/cos^2(3x-6)
Using the Pythagorean Identity for tangent:
tan^2(x) = sec^2(x) - 1
We can rewrite the equation as:
sec^2(3x-6) - 1 = (1 - cos^2(3x-6))/cos^2(3x-6)
Now, rearranging the equation and solving for cos^2(3x-6), we get:
cos^2(3x-6) = 1 / (1 + sec^2(3x-6))
Finally, substitute this expression for cos^2(3x-6) back into the original equation:
y = -2 + 7 * csc(3x-6)
= -2 + 7 * (1 / sin(3x-6))
= -2 + 7 * (1 / sqrt(1 / (1 + sec^2(3x-6))))
Therefore, we can write the equation y = -2 + 7csc(3x-6) in terms of secant as:
y = -2 + (7 / sqrt(1 + sec^2(3x-6)))