verify for sqrt(1-cosx)= sin/sqrt(1+cosx).
I multiplied the right side by sqrt(1-cosx) and got sqrt(1-cosx)= sin*sqrt(1-cosx), but I don't know what to do with the sin.
square both sides.
(1-cosx)=sin^2/(1+cosx)
multiple both sides by 1+cosx
1-cos^2 x = sin^2 x
sin^2x=sin^2x
Aren't you are only supposed to work on one side of the equation at a time? So you can't square or multiply both sides.
Ok ...
RS = sinx/√(1+cosx) * √(1-cosx)/√(1-cosx) , you did that
= sinx √(1-cosx) /(√( (1+cox)(1-cosx) )
= sinx √(1-cosx)/√(1 - cos^2 x)
= sinx √(1-cosx)/√sin^2 x
= sinx √(1-cosx)/sinx
= √(1-cosx)
= LS
To verify the equation sqrt(1-cosx) = sin/sqrt(1+cosx), we need to manipulate both sides of the equation in the same way.
Starting with the right side of the equation: sin/sqrt(1+cosx), we can rewrite it by rationalizing the denominator. To do this, we multiply both the numerator and denominator by sqrt(1+cosx):
(sin/sqrt(1+cosx)) * (sqrt(1+cosx)/sqrt(1+cosx))
= sin*sqrt(1+cosx) / (sqrt(1+cosx) * sqrt(1+cosx))
Now, we simplify the denominator:
sqrt(1+cosx) * sqrt(1+cosx)
= sqrt((1+cosx)*(1+cosx))
= sqrt(1 + 2cosx + cos^2(x))
Combining these simplifications, the right side becomes:
= sin*sqrt(1+cosx) / sqrt(1 + 2cosx + cos^2(x))
Next, we can simplify the left side of the equation, sqrt(1-cosx). We can square both sides to remove the square root:
(sqrt(1-cosx))^2 = (sin*sqrt(1+cosx) / sqrt(1 + 2cosx + cos^2(x)))^2
1 - cosx = (sin^2 * (1+cosx)) / (1 + 2cosx + cos^2(x))
Now, we need to simplify the right side further. We can distribute sin^2 to both terms in the numerator:
1 - cosx = (sin^2 + sin^2 * cosx) / (1 + 2cosx + cos^2(x))
Next, we can factor out sin^2(x) from the numerator:
1 - cosx = sin^2(x) * (1 + cosx) / (1 + 2cosx + cos^2(x))
Finally, we can simplify the right side:
1 - cosx = sin^2(x) / (1 + 2cosx + cos^2(x))
Now, we can see that the left side of the equation is equal to the right side:
1 - cosx = sin^2(x) / (1 + 2cosx + cos^2(x))
Therefore, this verifies that sqrt(1-cosx) = sin/sqrt(1+cosx).