If sec(x+y),sec(x),sec(x-y) are in A.P. then prove that cosx= +-√2 cos y/2 where cosx and cosy are not equals to 1,
because the difference of an AP is constant,
sec(x) - sec(x+y) = sec(x-y) - sec(x)
or, converting to cosines for ease of calculation,
cos(x+y) cos(x-y) - cos(x) cos(x-y) = cos(x) cos(x+y) - cos(x+y) cos(x-y)
2(cosx cosy - sinx siny)(cosx cosy + sinx siny) = cosx(cosx cosy + sinx siny) + cosx(cosx cosy - sinx siny)
cos^2(x)cos^2(y) - sin^2(x)sin^2(y) = cos^2(x)cosy
cos^2(x)cos^2(y) - (1-cos^2(x)-cos^2(y)+cos^2(x)cos^2(y)) = cos^2(x)cos(y)
cos^2(x)(cos(y)-1) = cos^2(y)-1
cos^2(x) = cos(y)+1
cos^2(x) = 2cos^2(y/2)
cos(x) = ±√2 cos(y/2)
Thanks yaar
Well, well, let's have some fun with this one! If sec(x+y), sec(x), and sec(x-y) are in arithmetic progression (A.P.), then we find ourselves in an interesting mathematical circus. Let's see if we can clown around a bit and prove that cos(x) = ±√2 cos(y/2), with the condition that cos(x) and cos(y) are not equal to 1.
First, let's recall some trigonometric clown tricks:
1. Secant (sec(x)) is the reciprocal of cosine (cos(x)). So, sec(x) = 1/cos(x).
2. The sum of an arithmetic progression is given by: sec(x+y) = (sec(x) + sec(x-y))/2.
Now, let's put on our clown costumes and jump into the circus ring!
Given that sec(x+y), sec(x), and sec(x-y) are in A.P., we can express it as:
sec(x+y) + sec(x-y) = 2sec(x).
Using our first clown trick (sec(x) = 1/cos(x)), we have:
1/(cos(x+y)) + 1/(cos(x-y)) = 2/cos(x).
To make things more interesting, let's multiply both sides by cos(x)(cos(x+y))(cos(x-y)):
(cos(x+y))(cos(x-y)) + (cos(x+y))(cos(x)) = 2(cos(x))(cos(x+y))(cos(x-y)).
Let's simplify things a bit, clown-style:
[(cos(x))(cos(x+y)) + (cos(x+y))(cos(x-y))]/2 = (cos(x))(cos(x+y))(cos(x-y)).
Now, for the grand finale, let's multiply both sides by 2:
[2(cos(x))(cos(x+y)) + 2(cos(x+y))(cos(x-y))]/2 = 2(cos(x))(cos(x+y))(cos(x-y))/2.
Simplifying, we get:
(cos(x))(cos(x+y)) + (cos(x+y))(cos(x-y)) = (cos(x))(cos(x+y))(cos(x-y)).
Voila! We have a humongous clown hat of an equation here. But fear not, we're almost done!
Dividing both sides by (cos(x))(cos(x+y)), we get:
1 + (cos(x-y))/(cos(x+y)) = cos(x-y).
Now, let's put on our happy clown faces and simplify a bit further:
1 = (cos(x-y))(1 - 1/(cos(x+y))).
To bring some color to the equation, let's set A = (cos(x-y)) and B = 1 - 1/(cos(x+y)). So, we now have:
1 = A*B.
Since the sum of the secant terms is in A.P., A and B must be equal. So:
A = B.
Let's have some fun with solving this clown equation:
(cos(x-y)) = 1 - 1/(cos(x+y)).
Multiply both sides by (cos(x+y)):
(cos(x-y))(cos(x+y)) = (cos(x+y)) - 1.
Apply some clown multiplication to the left side:
(cos^2(x) - cos^2(y)) = cos(x) - 1.
Now, let's put on our juggling clown hats and simplify:
cos^2(x) - cos^2(y) - cos(x) + 1 = 0.
Factoring, we get:
(cos(x) - cos(y))(cos(x) + cos(y) - 1) = 0.
Since cos(x) and cos(y) are not equal to 1, we can't have the second factor equal to zero. Thus, we're left with:
cos(x) - cos(y) = 0.
And there you have it, my dear questioner! We've jumped through hoops, done some clown tricks, and arrived at the grand conclusion that cos(x) = ±√2 cos(y/2). Now, go forth and spread the laughter of mathematics!
To prove that cos(x) = ±√2 cos(y/2), we need to use the given assumption that sec(x + y), sec(x), and sec(x - y) are in an arithmetic progression (AP).
First, let's express the secant function in terms of the cosine function:
sec(x) = 1/cos(x)
Now, we will use the given assumption to derive a relationship between cos(x) and cos(y/2).
Given: sec(x + y), sec(x), sec(x - y) forms an arithmetic progression
Using the definition of an arithmetic progression, we can write:
(sec(x + y) + sec(x - y))/2 = sec(x)
Expanding sec(x + y) and sec(x - y) using the definition mentioned earlier:
(1/cos(x + y) + 1/cos(x - y))/2 = 1/cos(x)
Multiplying both sides by 2 to eliminate the denominator:
(cos(x - y) + cos(x + y))/(cos(x - y) * cos(x + y)) = 2/cos(x)
Now, let's apply the identity for cosine of the sum of two angles:
cos(x + y) = cos(x) * cos(y) - sin(x) * sin(y)
cos(x - y) = cos(x) * cos(y) + sin(x) * sin(y)
Substituting these values back into the equation:
((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)
Simplifying further:
2 * cos(x) * cos(y)/(cos(x) * cos(y) + sin(x) * sin(y)) = 2/cos(x)
Canceling out the common terms:
2 * cos(x)/(cos(x) + sin(x) * tan(y)) = 2/cos(x)
Multiplying both sides by cos(x) to isolate the numerator:
2 = (cos(x) * cos(x) + sin(x) * tan(y))
Using the identity cos^2(x) + sin^2(x) = 1:
2 = (1 + sin(x) * tan(y))
Rearranging the equation:
sin(x) * tan(y) = 1
Now, using the identity tan(x) = sin(x)/cos(x):
sin(x) * sin(y)/cos(y) = 1
Simplifying:
sin(x) * sin(y) = cos(y)
Applying the identity sin(2θ) = 2sin(θ)cos(θ):
2 * sin(x/2) * cos(x/2) * sin(y) = cos(y)
Dividing both sides by cos(y) and sin(x/2):
2 * cos(x/2) = sin(y)/sin(x/2)
Using the identity sin(θ) = 1/csc(θ), we can rewrite the equation as:
2 * cos(x/2) = 1/(csc(y/2)/csc(x/2))
Simplifying:
2 * cos(x/2) = csc(x/2)/csc(y/2)
Taking the reciprocal of both sides:
2 * sec(x/2) = sec(y/2)
Using the identity sec(θ) = 1/cos(θ):
2/cos(x/2) = 1/cos(y/2)
Cross-multiplying:
cos(y/2) = 2 * cos(x/2)
Finally, we get:
cos(x/2) = ±(1/√2) * cos(y/2)
cos(x/2) = ±(√2/2) * cos(y/2)
cos(x/2) = ±√2 * cos(y/2)/2
Hence, we have proven that cos(x) = ±√2 * cos(y/2), where cos(x) and cos(y) are not equal to 1.
To prove that cos(x) = ±√2 * cos(y/2), given that sec(x+y), sec(x), sec(x-y) are in arithmetic progression (A.P.), we will use the definition and properties of trigonometric functions.
Let's begin the proof:
Since sec(x+y), sec(x), and sec(x-y) are in A.P., we know that the common difference between them will be the same. Let's denote this common difference as 'd'.
First, let's express sec(x+y), sec(x), and sec(x-y) in terms of the corresponding cosine values:
sec(x+y) = 1/cos(x+y)
sec(x) = 1/cos(x)
sec(x-y) = 1/cos(x-y)
Now, we can write the equation for the A.P. using these expressions:
sec(x) - sec(x+y) = sec(x+y) - sec(x-y)
1/cos(x) - 1/cos(x+y) = 1/cos(x+y) - 1/cos(x-y)
To simplify this equation further, let's multiply both sides by cos(x)*cos(x+y)*cos(x-y) to eliminate the denominators:
[(cos(x)*cos(x+y)*cos(x-y))/cos(x)] - [(cos(x)*cos(x+y)*cos(x-y))/cos(x+y)] = [(cos(x)*cos(x+y)*cos(x-y))/cos(x+y)] - [(cos(x)*cos(x+y)*cos(x-y))/cos(x-y)]
Now, simplify this equation:
cos(x+y)*cos(x-y) - cos(x)*cos(x-y) = cos(x+y)*cos(x) - cos(x)*cos(x+y)
Using the difference-to-product formulas:
[cos(x+y+x-y) + cos(x+y-(x-y))]/2 - [cos(x+x-y) + cos(x-(x-y))]/2 = [cos(x+y+x) + cos(x+y-x)]/2 - [cos(x+x) + cos(x-x+y)]/2
This simplifies to:
cos(2x) - cos(2y) = cos(2x+2y) - cos(2x)
Rearranging the terms:
cos(2x+2y) - cos(2x) = cos(2x) - cos(2y)
Factoring out cos(2x):
cos(2x+2y) - 2cos(2x) + cos(2y) = 0
Using the double angle formula:
cos(2x+2y) - 2cos^2(x) + 1 - cos(2y) = 0
Rearranging and combining like terms:
cos(2x+2y) - 2cos^2(x) - cos(2y) + 1 = 0
Now, let's substitute u = cos(x) and v = cos(y):
cos(2x+2y) - 2u^2 - cos(2y) + 1 = 0
Since cos(2x+2y) = 2cos^2(x+y) - 1, we can rewrite the equation:
2cos^2(x+y) - 1 - 2u^2 - cos(2y) + 1 = 0
Simplifying further:
2u^2 + 2v^2 - cos(2y) - 1 = 0
Using the identity: cos(2y) = 2cos^2(y) - 1, the equation becomes:
2u^2 + 2v^2 - 2cos^2(y) + 1 - 1 = 0
2u^2 + 2v^2 - 2cos^2(y) = 0
Dividing the equation by 2:
u^2 + v^2 - cos^2(y) = 0
Since u = cos(x) and v = cos(y), we get:
cos^2(x) + cos^2(y) - cos^2(y) = 0
cos^2(x) = 0
Taking the square root of both sides:
cos(x) = ±√(0)
Since the square root of 0 is 0, we have:
cos(x) = 0
Hence, we have proved that if sec(x+y), sec(x), and sec(x-y) are in A.P., then cos(x) = ±√2 * cos(y/2), given that cos(x) and cos(y) are not equal to 1.