Find all x such that 0 <= x <= pi/2 and sin^7 x + cos^7 x = 1. (Be sure to show that no other values of x satisfy the equation.)
Well, I know sin^2 x + cos^2 x = 1, and that kind of looks like sin^7 x + cos^7 x = 1. I'm unsure on how to proceed. Help?
let's look at
y = (sinx)^11 + (cosx)^11 , where x is a 1st quadrant angle, so both sinx and cosx would be positive
dy/dx = 11(sinx)^10 (cosx) + 11(cosx)^10 (-sinx)
= 0 for a max or min
11 (sinx)^10 cosx = 11 (cosx)^10 (sinx)
(sinx)^10 / cosx ^10 = sinx/cosx
(tanx)^10 = tanx
(tanx)^10 - tanx = 0
tanx( (tanx)^9 - 1) = 0
so tanx = 0 or tanx ^9 = 1
x = 0
or
x = π/4 which is 45°
if x=0 , then y = (sin 0)^11 + cos 0 ^ 11
= 0 + 1 = 1
ahh, that is one solution
if x = π/4
sin π/4 ^11 + cos π/4 ^11 = appr .044 ≠ 1
so our max must be 1 when x = 0
and our min must be appr .044 , when x = π/4
time to look at what Wolfram says about this
http://www.wolframalpha.com/input/?i=solve+%28sinx%29%5E11+%2B+%28cosx%29%5E11+%3D+1
interesting, notice that
x = 0 works
so does x = π/2 (90°) since sin π/2 = 1 but cos π/2 = 0
so for your given domain:
x = 0 or x = π/2 are the only two solutions.
Thanks a lot for your answer sir!
To solve the equation sin^7 x + cos^7 x = 1 for 0 <= x <= pi/2, we can use the fact that sin^2 x + cos^2 x = 1.
Let's start by rewriting the equation using the binomial expansion for powers of a binomial:
(sin^2 x + cos^2 x)(sin^5 x - sin^4 x cos x + sin^3 x cos^2 x - sin^2 x cos^3 x + sin x cos^4 x - cos^5 x) = 1
Since sin^2 x + cos^2 x = 1, we have:
sin^5 x - sin^4 x cos x + sin^3 x cos^2 x - sin^2 x cos^3 x + sin x cos^4 x - cos^5 x = 1
Next, we can rearrange the equation and factor out sin x and cos x:
sin^5 x(1 - cos x) + sin^3 x cos^2 x(1 - cos x) + sin x cos^4 x(1 - cos x) = cos^5 x
Now, notice that sin^5 x(1 - cos x), sin^3 x cos^2 x(1 - cos x), sin x cos^4 x(1 - cos x), and cos^5 x are all nonnegative. So, for the equation to hold, each term must be equal to zero:
sin^5 x(1 - cos x) = 0
sin^3 x cos^2 x(1 - cos x) = 0
sin x cos^4 x(1 - cos x) = 0
cos^5 x = 0
Let's consider each term separately:
1) sin^5 x(1 - cos x) = 0
This holds when either sin^5 x = 0 or (1 - cos x) = 0.
When sin^5 x = 0, we have sin x = 0. This gives us solutions x = 0 and x = pi (since sin x has period 2pi).
When (1 - cos x) = 0, we have cos x = 1. This gives us a solution x = 0.
2) sin^3 x cos^2 x(1 - cos x) = 0
This holds when either sin^3 x = 0, cos^2 x = 0, or (1 - cos x) = 0.
When sin^3 x = 0, we have sin x = 0. This gives us solutions x = 0 and x = pi.
When cos^2 x = 0, we have cos x = 0. This gives us a solution x = pi/2.
When (1 - cos x) = 0, we have cos x = 1. This gives us a solution x = 0.
3) sin x cos^4 x(1 - cos x) = 0
This holds when either sin x = 0, cos^4 x = 0, or (1 - cos x) = 0.
When sin x = 0, we have solutions x = 0 and x = pi.
When cos^4 x = 0, we have cos x = 0. This gives us a solution x = pi/2.
When (1 - cos x) = 0, we have cos x = 1. This gives us a solution x = 0.
4) cos^5 x = 0
This holds when cos x = 0. This gives us a solution x = pi/2.
Combining all the solutions we found, we have x = 0, x = pi/2, and x = pi.
Therefore, the values of x that satisfy the equation sin^7 x + cos^7 x = 1 for 0 <= x <= pi/2 are x = 0, x = pi/2, and x = pi. And no other values of x satisfy the equation.