PlEASE help me with this!! I need it as soon as possible.......
cos(x-(pi/4))+sin(x-(pi/4))=1
Find the answer from 0 to 2pi.
When I slove this, I got
cosx+sinx=(1/ square root of 2).
then I don't know what to do the rest. Please help me.
if you get to cosx+sinx=1/square root of 2, well we are doing identities in my pre-cal class and cosx+sinx always equals 1, so it would be 1/square root of 2.
Sorry, cosx+ sinx is not equal to 1 (from the identites). But, it was like this:
(sinx)^2+ (cosx)^2= 1
Can Anybody answer this question for me???? Please???
PlEASE help me with this!! I need it as soon as possible.......
cos(x-(pi/4))+sin(x-(pi/4))=1
Find the answer from 0 to 2pi
square both side:
(cosx+sinx)^x=(1/square root of 2)
then solve: it will work.
To solve the equation cos(x - (π/4)) + sin(x - (π/4)) = 1, you first need to simplify the equation.
1. Start by using the identity for the sine and cosine of the difference of two angles:
cos(A - B) = cosA * cosB + sinA * sinB
sin(A - B) = sinA * cosB - cosA * sinB
2. Apply these identities to the given equation:
cos(x - (π/4)) + sin(x - (π/4))
= cos(x) * cos(π/4) + sin(x) * sin(π/4) + sin(x) * cos(π/4) - cos(x) * sin(π/4)
= (cos(x) + sin(x)) * cos(π/4) + (sin(x) - cos(x)) * sin(π/4)
3. Simplify cos(π/4) and sin(π/4):
cos(π/4) = √2/2
sin(π/4) = √2/2
4. Substitute these values back into the equation:
(cos(x) + sin(x)) * (√2/2) + (sin(x) - cos(x)) * (√2/2) = 1
5. Rearrange the equation by combining like terms:
(√2/2) * cos(x) + (√2/2) * sin(x) - (√2/2) * cos(x) - (√2/2) * sin(x) = 1
Simplifying further, you get:
0 = 1
6. Since the equation simplifies to an invalid statement (0 = 1), there are no solutions to the original equation cos(x - (π/4)) + sin(x - (π/4)) = 1 within the interval [0, 2π].