Let x and y be real numbers, with 0 < x < pi/2 and pi/2 < y < pi. sin y = 4/5 and 11sin x + 5cos (y - x) = 3. Find cos y and sin 2x.

Answer 1

Regarding x and y let's take the requested equation and break it down into its simplest forms to use the given information...

x in Q1 and y in Q2

sin(y) = 4/5 --> π/2 < y < π

11sin(x) + 5cos(y-x) = 3

Using the angle difference equation for cosine function --> cos(a - b) = cos(a)cos(b) - sin(a)sin(b)

11sin(x) + 5*[cos(y)*cos(x) - sin(y)*sin(x)] = 3

We know sin(y) and we can figure out cos(y) using the Pythagorean Theorem --> cos²(a) + sin²(a) = 1

cos(y) = +/- sqrt(1 - (4/5)²) = +/- 3/5

Since y is in Q2 we need to choose the negative value from the sqrt since cos is negative in the Q2.

Therefore cos(y) = -3/5

Our equation now looks like...

11sin(x) + 5(-3/5)*cos(x) - 5(4/5)*sin(x) = 3

11sin(x) - 3cos(x) - 5sin(x) = 3

6sin(x) - 3cos(x) = 3

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

Now we work to compute sin(2x)

sin(2x) = 2sin(x)cos(x)

If we can compute sin(x) and cos(x) we can compute sin(2x). The equation we have above relates sin(x) to cos(x) via --> 2sin(x) - cos(x) = 1

Square both sides --> 4sin²(x) + cos²(x) - 4sin(x)cos(x) = 1

Rearrange and simplify where possible...

3sin²(x) + sin²(x) + cos²(x) - 2sin(2x) = 1

3sin²(x) + 1 - 2sin(2x) = 1

3sin²(x) - 2sin(2x) = 0

3sin²(x) = 2sin(2x)

sin(2x) = 3/2*sin²(x)

If we know sin(x) we can now compute sin(2x)...

We also know sin(2x) = 2sin(x)cos(x) so we can solve for sin(x) in terms of sin(2x)...

2sin(x)cos(x) = 3/2*sin²(x)

2cos(x) = 3/2 * sin(x)

Now build the tangent function from (sin/cos) --> 4/3 = tan(x) and since x is in Q1, the tangent value should be positive (which it is)...

x = arctan(4/3) --> sin(2x) = sin(2*arctan(4/3)) = 0.96

Therefore the sin(2x) = 0.96