Sin - cos = - radical 2
I don't understand this. It wants you to make the right side equal to zero and solve
sinx - cosx = √2
square both sides and you have
sin^2x - 2sinx cosx + cos^2x = 2
-2sinx cos x = 1
sin 2x = -1
2x = 3pi/2
x = 3pi/4
Or, note that
sinx - cosx
= √2 (sinx * 1/√2 - cosx * 1/√2)
= √2 sin(x - pi/4)
so,
√2 sin(x-pi/4) = √2
sin(x-pi/4) = 1
x - pi/4 = pi/2
x = 3pi/4
I see that I missed the fact that it was -√2 on the right. Just make the fix and redo the calculations to find that x = 7pi/4
If - radical 2 is squared wouldn't it still turn out to be positive 2?
To solve the equation sin(x) - cos(x) = -√2, we need to rearrange it to have zero on one side.
First, let's add √2 to both sides of the equation:
sin(x) - cos(x) + √2 = 0
Now, we can use trigonometric identities to rewrite sin(x) and cos(x).
Recall that sin(x) = cos(π/2 - x).
Substituting that into the equation, we have:
cos(π/2 - x) - cos(x) + √2 = 0
Since cos(a) - cos(b) = -2sin((a+b)/2)sin((a-b)/2), we can use this identity to simplify the equation further:
-2sin((π/2 - x + x)/2)sin((π/2 - x - x)/2) + √2 = 0
Simplifying this expression gives us:
-2sin(π/4)sin(-x/2) + √2 = 0
Since sin(π/4) = 1/√2 and sin(-θ) = -sin(θ), we can simplify further:
-2 * (1/√2) * sin(-x/2) + √2 = 0
Now, simplifying the expression gives us:
-√2sin(-x/2) + √2 = 0
Since sin(-θ) = -sin(θ), the equation becomes:
√2sin(x/2) + √2 = 0
To solve for x, divide both sides by √2:
sin(x/2) + 1 = 0
Now, isolate sin(x/2):
sin(x/2) = -1
To solve this equation for x/2, we take the inverse sine (sin^(-1)) of both sides:
x/2 = sin^(-1)(-1)
The inverse sine of -1 is -π/2.
Thus, x/2 = -π/2
To find the value of x, multiply both sides by 2:
x = -π