prove sin(x+y)+sin(x-y)/cos(x+y)+cos(x-y)=tanx
Use trigonometric identities:
sin ( x + y ) = sin x cos y + cos x sin y
sin ( x - y ) = sin x cos y - cos x sin y
cos ( x + y ) = cos x cos y - sin x sin y
cos ( x - y ) = sin x sin y + cos x cos y
sin ( x + y ) + sin ( x - y ) =
sin x cos y + cos x sin y + sin x cos y - cos x sin y =
sin x cos y + sin x cos y + cos x sin y - cos x sin y =
2 sin x cos y
cos ( x + y ) + cos ( x - y ) =
cos x cos y - sin x sin y + sin x sin y + cos x cos y =
cos x cos y + cos x cos y - sin x sin y + sin x sin y =
2 cos x cos y
[ sin ( x + y ) + sin ( x - y ) ] / [ cos ( x + y ) + cos ( x - y ) ] =
2 sin x cos y / 2 cos x cos y =
sin x / cos x = tan x
Watch those brackets, you must have meant:
(sin(x+y)+sin(x-y))/(cos(x+y)+cos(x-y))=tanx
LS =
(sinxcosy + cosxsiny + sinxcosy - sinxcosy)/(cosxcosy - sinxsiny + cosxcosy + sinxsiny)
= 2sinxcosy/(2cosxcosy)
= (sinx/cosx)(cosy/cosy)
= tanx
= RS
To prove the given equation, we'll start with the left-hand side (LHS) and simplify it step-by-step to see if it simplifies to the right-hand side (RHS).
LHS: (sin(x + y) + sin(x - y)) / (cos(x + y) + cos(x - y))
Step 1: Expand the numerator
We can expand the numerator using the sum-to-product formulas for sine:
LHS: (sin(x)cos(y) + cos(x)sin(y) + sin(x)cos(y) - cos(x)sin(y)) / (cos(x + y) + cos(x - y))
Simplifying the numerator:
LHS: (2sin(x)cos(y)) / (cos(x + y) + cos(x - y))
Step 2: Expand the denominator
Similarly, we can expand the denominator using the sum-to-product formulas for cosine:
LHS: (2sin(x)cos(y)) / [2cos(x)cos(y)]
Canceling out the common factors of 2 and cos(y):
LHS: sin(x) / cos(x)
Step 3: Simplify sin(x) / cos(x)
The expression sin(x) / cos(x) is equivalent to the tangent of x, so we have:
LHS: tan(x)
Since the LHS simplifies to the RHS, we have proved the given equation:
(sin(x + y) + sin(x - y)) / (cos(x + y) + cos(x - y)) = tan(x)
To prove the trigonometric identity sin(x+y) + sin(x-y) / cos(x+y) + cos(x-y) = tan(x), we'll start by simplifying the expression on the left-hand side (LHS) using trigonometric identities.
First, we'll write the numerator of the expression (sin(x+y) + sin(x-y)) as a single term:
sin(x+y) + sin(x-y) = (sin x cos y + cos x sin y) + (sin x cos y - cos x sin y)
= sin x cos y + cos x sin y + sin x cos y - cos x sin y
= 2 sin x cos y
Next, we'll write the denominator of the expression (cos(x+y) + cos(x-y)) as a single term:
cos(x+y) + cos(x-y) = (cos x cos y - sin x sin y) + (cos x cos y + sin x sin y)
= cos x cos y - sin x sin y + cos x cos y + sin x sin y
= 2 cos x cos y
Now, we can simplify the expression:
LHS = (2 sin x cos y) / (2 cos x cos y)
Dividing both the numerator and denominator by 2, we get:
LHS = (sin x cos y) / (cos x cos y)
Now, we can simplify further by canceling out the common factors (cos y):
LHS = sin x / cos x
Finally, using the definition of tan x, we have:
LHS = tan x
Thus, we have proven that sin(x+y) + sin(x-y) / cos(x+y) + cos(x-y) = tan x.