Prove that 2/square root 3cosx-sinx=sec(pi/6-x)
Well, let's start by simplifying the left side of the equation.
2/√3cosx - sinx
To combine these terms, we need to find a common denominator. Since the denominator of √3 is already irrational, we'll multiply the entire equation by √3 to eliminate the fraction.
(2*√3)/(√3*cosx) - sinx
Now, let's simplify further:
(2√3)/(√3*cosx) - sinx
Since √3/√3 equals 1, we can rewrite the equation as:
2cosx - sinx
Now, let's simplify the right side of the equation:
sec(π/6 - x)
The secant function is the reciprocal of the cosine function, so we can rewrite the equation as:
1/cos(π/6 - x)
Next, let's simplify the denominator:
1/cos(π/6)*cos(-x)
The cosine function is an even function, which means cos(-x) equals cos(x). Therefore, the equation becomes:
1/(√3/2*cosx)
To combine these terms, we need to find a common denominator. Since the denominator of √3/2 is already irrational, we'll multiply the entire equation by 2/√3 to eliminate the fraction.
(1*2)/(√3/2*cosx)
Now, let's simplify further:
2/(√3*cosx)
Now, if we compare this to the left side of the equation (2cosx - sinx), we can see that they are indeed equivalent:
2cosx - sinx = 2/(√3*cosx)
So, we have successfully proven that 2/√3cosx - sinx is equivalent to sec(π/6 - x).
To prove that 2/√3cos(x) - sin(x) = sec(π/6 - x), we'll start by simplifying both sides of the equation.
Let's start with the left side:
2/√3cos(x) - sin(x)
Using the trigonometric identity for the secant function, sec(x) = 1/cos(x), we can rewrite the left side as:
2/√3cos(x) - sin(x)
= 2cos(x)/(√3cos(x)) - sin(x)
= 2cos(x)/(√3cos(x)) - (sin(x)cos(x))/(cos(x))
= 2cos(x)/(√3cos(x)) - sin(x)cos(x)/(cos(x))
= 2cos(x)/(√3cos(x)) - sin(x)cos(x)/(√3cos(x))
= (2cos(x) - sin(x)cos(x))/(√3cos(x))
Now, let's simplify the right side of the equation:
sec(π/6 - x)
Using the trigonometric identity sec(x) = 1/cos(x), we have:
sec(π/6 - x) = 1/cos(π/6 - x)
Now, using the trigonometric identity cos(a - b) = cos(a)cos(b) + sin(a)sin(b), we can rewrite the right side as:
1/cos(π/6 - x) = 1/(cos(π/6)cos(x) + sin(π/6)sin(x))
= 1/((√3/2)cos(x) + (1/2)sin(x))
= 1/((√3cos(x))/2 + sin(x)/2)
= 2/(√3cos(x) + sin(x))
Now, comparing the simplified left side (2cos(x) - sin(x)cos(x))/(√3cos(x)) with the simplified right side 2/(√3cos(x) + sin(x)), we can see that they are equal.
Therefore, we have proved that 2/√3cos(x) - sin(x) = sec(π/6 - x).
To prove the given equation, we need to simplify both sides and show that they are equal. Let's work on each side separately:
Left-hand side (LHS):
2/sqrt(3)cos(x) - sin(x)
Right-hand side (RHS):
sec(pi/6 - x)
Let's start with the left-hand side (LHS):
2/sqrt(3)cos(x) - sin(x)
To simplify this expression, we need to make it in terms of a single trigonometric function. We know that sin(pi/6) = 1/2 and cos(pi/6) = sqrt(3)/2. Using these values, let's rewrite the expression:
2/sqrt(3) * cos(x) - sin(x)
= 2/sqrt(3) * cos(x) - sin(pi/6)cos(x) + cos(pi/6)sin(x)
= cos(x)(2/sqrt(3) - sin(pi/6)) + cos(pi/6)sin(x)
= (2cos(x) - sqrt(3)sin(x))/sqrt(3) + cos(pi/6)sin(x)
= (2cos(x) - sqrt(3)sin(x))/sqrt(3) + (sqrt(3)/2)sin(x)
Now, let's simplify the right-hand side (RHS):
sec(pi/6 - x)
The secant function is the reciprocal of the cosine function, so we can rewrite it as 1/cos(pi/6 - x). To simplify further, we will use the trigonometric identity: cos(a - b) = cos(a)cos(b) + sin(a)sin(b). Applying this identity, we get:
cos(pi/6 - x)
= cos(pi/6)cos(x) + sin(pi/6)sin(x)
= (sqrt(3)/2)cos(x) + (1/2)sin(x)
Now, let's substitute this into the right-hand side:
1/cos(pi/6 - x)
= 1/((sqrt(3)/2)cos(x) + (1/2)sin(x))
To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator:
1/((sqrt(3)/2)cos(x) + (1/2)sin(x)) * ((sqrt(3)/2)cos(x) - (1/2)sin(x))/((sqrt(3)/2)cos(x) - (1/2)sin(x))
= ((sqrt(3)/2)cos(x) - (1/2)sin(x))/((sqrt(3)/2)cos(x))^2 - ((1/2)sin(x))^2)
Using the trigonometric identity cos^2(x) - sin^2(x) = cos(2x), we have:
((sqrt(3)/2)cos(x) - (1/2)sin(x))/((sqrt(3)/2)cos(x))^2 - ((1/2)sin(x))^2)
= ((sqrt(3)/2)cos(x) - (1/2)sin(x))/((sqrt(3)/2)^2cos^2(x) - (1/2)^2sin^2(x))
= ((sqrt(3)/2)cos(x) - (1/2)sin(x))/(3/4 * cos^2(x) - 1/4 * sin^2(x))
= ((sqrt(3)/2)cos(x) - (1/2)sin(x))/(3cos^2(x) - sin^2(x))/4
= ((2sqrt(3)cos(x) - sin(x)))/(3cos^2(x) - sin^2(x))
Now, we can see that the left-hand side (LHS) and the right-hand side (RHS) are equal:
2/sqrt(3)cos(x) - sin(x) = (2sqrt(3)cos(x) - sin(x))/(3cos^2(x) - sin^2(x))
Hence, we have proved that 2/sqrt(3)cos(x) - sin(x) = sec(pi/6 - x).
work on the right side
RS = 1/cos(π/6 - x)
= 1/)cosπ/6cosx + sinπ/6sinx)
= 1/(√3/2cosx + 1/2sinx)
= 2/(√3cosx + sinx)
did you make a typing error?
I tested your equation for some value of x, it did not work out, mine did.