Trig identities conjugates and common denominators?
Common denominator and conjugate in solving/proving trig identities, I was wondering if someone could explain.
For example: Common denominator: Sinx/CosX+Cosx/Sinx = Sin^2xCos^2x/CosxSinx was wondering how this is determined, well i see it but is there a rule to finding it.
and the conjugates for a function such as Sinx/1+cosx
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(sin x / cos x) + (cos x / sin x) = (sin²x + cos²x) / cos x sin x, NOT (sin x / cos x) + (cos x / sin x) = sin²x cos²x / cos x sin x
(sin x / cos x) + (cos x / sin x) =
The LCD is cos x * sin x = cos x sin x. Multiply as needed.
(sin x / cos x)(sin x / sin x) + (cos x / sin x)(cos x / cos x) =
(sin²x / cos x sin x) + (cos²x / cos x sin x) =
Now that the denominators are the same, you can rewrite it as a single fraction by combining the numerators and writing the result over the common denominator.
(sin²x + cos²x) / cos x sin x
~~~~~~~~~~~~~~~~~~~~~
If you're having trouble seeing this, replace sin x with a and cos x with b:
(a / b) + (b / a) =
(a / b)(a / a) + (b / a)(b / b) =
(a² / ab) + (b² / ab) =
(a² + b²) / ab
~~~~~~~~~~~~~~~~~~~~~
Remember that sin²x + cos²x = 1.
1 / cos x sin x =
If you wish, you can also translate to other trig identities:
Factor.
(1 / cos x)(1 / sin x) =
Remember that sec x = 1 / cos x
sec x(1 / sin x) =
Remember that csc x = 1 / sin x.
sec x csc x
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
sin x / (1+ cos x) =
Remember that the conjugate of (a + b) is (a - b). Multiply as needed.
[sin x / (1+ cos x)][(1 - cos x) / (1 - cos x)] =
[sin x(1 - cos x)] / [(1+ cos x)(1 - cos x)] =
Distribute.
[sin x(1) + sin x(-cos x)] / [(1+ cos x)(1 - cos x)] =
(sin x - sin x cos x) / [(1+ cos x)(1 - cos x)] =
Notice that the denominator is the factored form of the difference of two squares.
(sin x - sin x cos x) / [(1)² - (cos x)²] =
(sin x - sin x cos x) / (1 - cos²x) =
Remember that 1 - cos²x = sin²x.
(sin x - sin x cos x) / sin²x =
Factor the numerator.
sin x(1 - cos x) / sin²x =
Simplify.
1(1 - cos x) / sin x =
(1 - cos x) / sin x =
Distribute the denominator.
(1 / sin x) - (cos x / sin x) =
Remember that csc x = 1 / sin x.
csc x - (cos x / sin x) =
Remember that cos x / sin x = cot x.
csc x - cot x
There are no reducible entities in the equation, so multiply it out and see if what you get is true. If it is, then the equation is proven. cos A + sin Acos A + cos^2 A = 1 - sin A + cos A + sin A - sin^2 A + Sin Acos A cosA and sinAcosA on both sides cancel out, leaving cos^2A = 1 - sin^2A, or to put it another way, sin^2 A + cos^2 A = 1, which we know to be true as it is one of the Pythagorean identities*. Therefore, (1 + sinA + cosA )/(1 - sinA + cosA ) = ( 1 + sinA )/cosA *sin^2 A + cos^2 A = a^2/h^2 + b^2/h^2 = (a^2 + b^2)/*h^2 = h^2/h^2 = 1
Common denominators
L.S.
= sin(x)/cos(x) + cos(x)/sin(x)
= [sin^2(x)]/[sin(x)cos(x)] + [cos^2(x)] / [sin(x)cos(x)]
= [sin^2(x) + cos^2(x) / [sin(x)cos(x)]
= 1 / [sin(x)cos(x)]
R.S.
= [sin^2(x)*cos^2(x)] / [cos(x)*sin(x)]
= sin(x)*cos(x)
==> L.S = 1 / R.S
==> L.S. <> R.S.