verify this trigonometric identity
cos 'a' - cos 'b' / sin 'a' + sin 'b'
+
sin 'a' - sin 'b' / cos 'a' + cos 'b'
= 0
please help me with this! i don't know how to do it!
I am sure that you meant:
(cos 'a' - cos 'b') / (sin 'a' + sin 'b')
+
(sin 'a' - sin 'b') / (cos 'a' + cos 'b')
= 0
the LCD is (sina + sinb)(cosa + cosb)
LS = ( (cosa - cosb)(cosa + cosb) + sina - sinb)(sina - sinb) ) / ( (sina + sinb)(cosa + cosb) )
= (cos^2 a - cos^2 b + sin^2 a - sin^2 b) / ( (sina + sinb)(cosa + cosb) )
= ( cos^2 a + sin^2 a - (cos^2 ba + sin^2 b) / ( (sina + sinb)(cosa + cosb) )
= (1-1)/( (sina + sinb)(cosa + cosb) )
= 0/( (sina + sinb)(cosa + cosb) )
= 0
= RS
For clutter reduction, let's let
x = (a+b)/2
y = (a-b)/2
using the sum-to-product formulas,
cosa-cosb = -2 sinx siny
sina+sinb = 2 sinx cosy
sina-sinb = 2 cosx siny
cosa+cosb = 2 cosx cosy
Now we just have
-tany + tany = 0
seems legit to me.
To verify the given trigonometric identity, we need to simplify both sides of the equation separately and see if they are equal.
Let's start with the left side of the equation:
cos 'a' - cos 'b' / sin 'a' + sin 'b'
+
sin 'a' - sin 'b' / cos 'a' + cos 'b'
First, let's simplify the fractions using the common denominator sin 'a' + sin 'b' / (sin 'a' + sin 'b') and cos 'a' + cos 'b' / (cos 'a' + cos 'b'):
[(cos 'a' - cos 'b') * (sin 'a' + sin 'b')] / [(sin 'a' + sin 'b') * (cos 'a' + cos 'b')] + [(sin 'a' - sin 'b') * (cos 'a' + cos 'b')] / [(sin 'a' + sin 'b') * (cos 'a' + cos 'b')]
Now, let's simplify each term:
(cos 'a' * sin 'a' + cos 'a' * sin 'b' - cos 'b' * sin 'a' - cos 'b' * sin 'b') / [(sin 'a' + sin 'b') * (cos 'a' + cos 'b')]
+
(sin 'a' * cos 'a' + sin 'a' * cos 'b' - sin 'b' * cos 'a' - sin 'b' * cos 'b') / [(sin 'a' + sin 'b') * (cos 'a' + cos 'b')]
Now, let's combine like terms:
(cos 'a' * sin 'a' - cos 'b' * sin 'a' + cos 'a' * sin 'b' - cos 'b' * sin 'b' + sin 'a' * cos 'a' - sin 'b' * cos 'a' + sin 'a' * cos 'b' - sin 'b' * cos 'b']) / [(sin 'a' + sin 'b') * (cos 'a' + cos 'b')]
Now, notice that each term can be grouped into either a sin 'a' or sin 'b' factor and a cos 'a' or cos 'b' factor:
[(cos 'a' * sin 'a' - sin 'a' * cos 'a') + (cos 'a' * sin 'b' - sin 'b' * cos 'a')] + [(cos 'b' * sin 'a' - sin 'a' * cos 'b') + (cos 'b' * sin 'b' - sin 'b' * cos 'b')] / [(sin 'a' + sin 'b') * (cos 'a' + cos 'b')]
Using the fact that cos 'a' * sin 'a' - sin 'a' * cos 'a' is 0 and cos 'b' * sin 'b' - sin 'b' * cos 'b' is 0, we can simplify further:
[(cos 'a' * sin 'b' - sin 'b' * cos 'a')] + [(cos 'b' * sin 'a' - sin 'a' * cos 'b')] / [(sin 'a' + sin 'b') * (cos 'a' + cos 'b')]
Now, notice that cos 'a' * sin 'b' - sin 'b' * cos 'a' is the same as sin ('b - a') using the trigonometric identity sin (x - y) = sin x * cos y - cos x * sin y. Similarly, cos 'b' * sin 'a' - sin 'a' * cos 'b' is the same as sin ('a - b').
Therefore, the left side of the equation simplifies to:
sin ('b - a') + sin ('a - b') / (sin 'a' + sin 'b') * (cos 'a' + cos 'b')
Now, let's simplify the right side of the equation:
0
Since the right side is already 0, we have verified the given trigonometric identity.