# if tan 2x= -b/a, pi/2 <=2x<=pi, then determine an expression for sinx cosx in terms of a and b.

## the two legs are a and b, and the hypotenuse is √(a^2+b^2). So, assuming a,b > 0,

So, in QII,
sin2x = b/√(a^2+b^2)
cos2x = -a/√(a^2+b^2)

sin^2(x) = (1-cos2x)/2 = (1+a/√(a^2+b^2))/2
= (a+√(a^2+b^2)) / 2√(a^2+b^2)

and similarly for cosx

## To find an expression for sinx cosx in terms of a and b, we'll use the double angle formula for tangent:

tan 2x = 2tanx / (1 - tan^2x)

Given that tan 2x = -b/a, we can substitute this expression into the double angle formula:

(-b/a) = 2tanx / (1 - tan^2x)

Next, we'll solve this equation for tanx:

(-b/a)(1 - tan^2x) = 2tanx

-b + (b/a)tan^2x = 2tanx

(b/a)tan^2x + 2tanx - b = 0

Now, we can solve this quadratic equation for tanx using the quadratic formula:

tanx = (-b/a ± √((b/a)^2 - 4(b/a)) / (2(b/a))

After simplifying, we get:

tanx = (-b ± √(b^2 - 4ab)) / (2a)

Since we know that pi/2 ≤ 2x ≤ pi, we can determine the range of values for x. Since tan is negative in the second quadrant (pi/2 < x < pi), we choose the negative value:

tanx = (-b - √(b^2 - 4ab)) / (2a)

Now we can use the trigonometric identity for sine and cosine:

sin^2x + cos^2x = 1

Squaring both sides gives us:

(sin^2x) + 2sinxcosx + (cos^2x) = 1

Rearranging the terms, we get:

2sinxcosx = 1 - (sin^2x) - (cos^2x)

Using the Pythagorean identity sin^2x = 1 - cos^2x, we can substitute this expression and simplify:

2sinxcosx = 1 - (1 - cos^2x) - (cos^2x)

= 1 - 1 + cos^2x - cos^2x

= 0

Therefore, the expression for sinx cosx in terms of a and b is 0.

## To find an expression for sin(x) cos(x) in terms of a and b, we can manipulate the given equation and use trigonometric identities.

Given:
tan(2x) = -b/a
pi/2 <= 2x <= pi

First, we'll use the double-angle identity for tangent:
tan(2x) = (2tan(x))/(1 - tan^2(x))

Substituting the given value:
(-b/a) = (2tan(x))/(1 - tan^2(x))

Next, simplify the equation:
(-b/a)(1 - tan^2(x)) = 2tan(x)
-b + (b/a)tan^2(x) = 2tan(x)

Rearranging the terms:
(b/a)tan^2(x) + 2tan(x) - b = 0

Since the equation is quadratic in nature, we'll use the quadratic formula:
tan(x) = [-2 ± √(4 + 4(b/a)(b))]/(2(b/a))

Simplifying further:
tan(x) = [-1 ± √(1 + (b^2/a^2))]/(b/a)

After finding the value of tan(x), we can use the trigonometric identity sin^2(x) + cos^2(x) = 1. Let's solve for cos(x)^2 first:

sin^2(x) + cos^2(x) = 1
=> cos^2(x) = 1 - sin^2(x)

Now, we can substitute tan(x) = sin(x)/cos(x) using:
tan^2(x) = sin^2(x) / cos^2(x)

Substituting these equations into the previous expression for tan(x):

(b/a)(sin^2(x) / cos^2(x)) + 2(sin(x) / cos(x)) - b = 0

Multiplying through by cos^2(x) to eliminate the denominator:

b*sin^2(x) + 2sin(x)*cos(x) - b*cos^2(x) = 0

Using the identity cos^2(x) = 1 - sin^2(x):

b*sin^2(x) + 2sin(x)*cos(x) - b + b*sin^2(x) = 0

Combining like terms:

2sin(x)*cos(x) + 2b*sin^2(x) - b = 0

Finally, factoring out a common factor:
2sin(x)*(cos(x) + b*sin(x)) - b = 0

Therefore, the expression for sin(x)*cos(x) in terms of a and b is:
(sin(x))*(cos(x) + b*sin(x))/2a