sinxsq + sinxcosx=0?

Question

and how to slove sinxsq(pi/8)

Answer 1

I assume you want to solve

sin^2(x) + sin(x)cos(x) = 0

I'm going to assume an interval of 0 <= x < 2pi

To solve this, we factor out sin(x).

sin(x) [sin(x) + cos(x)] = 0

Now, we equate each factor to 0. That is,
sin(x) = 0, sin(x) + cos(x) = 0

If sin(x) = 0, this occurs at the two points x = {0, pi}.

If sin(x) + cos(x) = 0, then
sin(x) = -cos(x), and dividing both sides by cos(x), we have
sin(x)/cos(x) = -1. Therefore,
tan(x) = -1, and this is true at
x = {3pi/4, 7pi/4}

Therefore, our solutions are x = {0, pi, 3pi/4, 7pi/4}

b) sin^2(pi/8)

Use the half angle formula, which goes as follows:

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

Therefore,

sin^2(pi/8) = (1/2) (1 - cos(2*pi/8))
sin^2(pi/8) = (1/2) (1 - cos(2pi/8))
sin^2(pi/8) = (1/2) (1 - cos(pi/4))

Note that pi/4 is actually one of our known values on the unit circle. cos(pi/4) = sqrt(2)/2, so we have

sin^2(pi/8) = (1/2) (1 - sqrt(2)/2)
sin^2(pi/8) = (1/2) (2 - sqrt(2))/2
sin^2(pi/8) = [2 - sqrt(2)]/4

Answer 2

I guess that you mean:

sin^2 x + sin x cos x = 0

sin x(sin x + cos x) = 0

So, sin x = 0 or sin x + cos x = 0

If sin x = 0, then x = k pi, k an integer.

If sin x + cos x = 0, then sin x = - cos x (1)

If cos x = 0, then x = k (pi/2), so sin x = 1, then sin x + cos x isnt equals to 0

If cos x isnt 0, then I can divide (1) by cos x and I get

sinx / cos x = tan x = -1. So, x = -pi/4 + kpi, k an integer


To slove sin^2 (pi/8), I cant find a clever method, so, as a friend of mine says, lets use ignorance and brute force

sin (2x) = 2sin x cos x

sin (pi/4) = 2 sin (pi/8) cos (pi/8) = 1/V2

sin (pi/8) cos (pi/8) = 1/(2V2) = 1/[V(2)^2 V2] = 1/V8

sin^2 (pi/8) cos^2 (pi/8) = 1/8

sin^2 (pi/8) [1-sin^2 (pi/8)] = 1/8

sin^2 (pi/8) - sin^4 (pi/8) = 1/8

Use sin^2 = z and solve for z.


Ana

Answer 3

a) sin^2 x + sin x cos x = 0
=> sin x (sin x + cos x) = 0
=> sin x = 0 or sin x = - cos x
=> sin x = 0 or tan x = -1
=> x = nπ or (n + 3/4)π for any n ∈ Z.

b) sin^2 (π/8)
Note from double-angle formulae
cos(2x) = cos^2 x - sin^2 x = 1 - 2 sin^2 x
So sin^2 x = (1 - cos 2x) / 2.
So sin^2 (π/8) = (1 - cos (π/4)) / 2
= (1 - 1/√2) / 2
= (2 - √2) / 4.