prove identity
Sin2x - sinx/cosx + cos2x= sinx/cosx + 1
You typed it incorrectly, you must mean:
(sin(2x) - sinx)/(cosx + cos(2x)) = sinx/(cosx + 1)
LS = (2sinxcosx - sinx)/(cosx + 2cos^2 x - 1)
= sinx(2cosx -1)/( (2cosx - 1)(cosx+1) )
= sinx/(cosx + 1)
= RS
To prove the identity:
Sin(2x) - sin(x)/cos(x) + cos(2x) = sin(x)/cos(x) + 1
Let's simplify and work with each side of the equation separately:
Starting with the left side of the equation:
Sin(2x) - sin(x)/cos(x) + cos(2x)
Using the double angle formula for sin(2x) = 2sin(x)cos(x) and cos(2x) = cos²(x) - sin²(x):
2sin(x)cos(x) - sin(x)/cos(x) + cos²(x) - sin²(x)
Combining like terms:
(sin²(x) - sin²(x)) / cos(x) + cos²(x) + 2sin(x)cos(x) / cos(x)
Simplifying further:
0/cos(x) + cos²(x) + 2sin(x)
Since 0/cos(x) is equal to 0, the expression becomes:
cos²(x) + 2sin(x)
Now let's simplify the right side of the equation:
sin(x)/cos(x) + 1
We can combine the two fractions:
(sin(x) + cos(x))/cos(x)
Now, we need to rewrite (cos²(x) + 2sin(x)) in terms of cos(x).
Using the identity sin²(x) = 1 - cos²(x):
cos²(x) + 2sin(x) = (1 - sin²(x)) + 2sin(x)
Expanding and rearranging terms:
1 - sin²(x) + 2sin(x)
Now, using the Pythagorean identity sin²(x) + cos²(x) = 1:
1 - sin²(x) + 2sin(x) = cos²(x) + 2sin(x)
Now both sides of the equation are equal:
cos²(x) + 2sin(x) = cos²(x) + 2sin(x)
Therefore, we have successfully proved the identity:
Sin(2x) - sin(x)/cos(x) + cos(2x) = sin(x)/cos(x) + 1