Question 551282: I've been having some problems trying to solve this,
sin(x+pi/4) + sin(x-pi/4) = 1 for the interval [0,2pi]
I don't really even know where to start, I think it has something to do with co-function identities, but I don't know how to translate that since it's pi/4 and not pi/2...
Help is greatly appreciated, Thank you for your time
sin(x+pi/4) + sin(x-pi/4) = 1 for the interval [0,2pi]
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! I've been having some problems trying to solve this,
sin(x+pi/4) + sin(x-pi/4) = 1 for the interval [0,2pi]
**
sin(x+pi/4) + sin(x-pi/4) = 1
Addition formulas for sin:
sin(s+t)=sin s cos t + cos s sin t
sin(s-t)=sin s cos t - cos s sin t
..
sin(x+π/4)=sin x cos π/4 + cos x sin π/4
=sin x*√2/2+cos x*√2/2=√2/2 sin x + √2/2 cos x
..
sin(x-π/4)=sin x cos π/4 - cos x sin π/4
=sin x*√2/2-cos x*√2/2=√2/2 sin x - √2/2 cos x
..
√2/2 sin x + √2/2 cos x + √2/2 sin x - √2/2 cos x=1
√2 sin x=1
sin x=1/√2
x=π/4
|
|
|
