Question

IF cos●+cos○=sin●+sin○=0then show that ,a) cos2●+cos2○=2cos(pi+●+○) b) sin2●+sin2○=2sin(pi+●+○). where ● is theta and ○ is sai.​

Answer 1

Let me consider theta as 'a' and chay as 'b'.

Now ,

cos(a) + cos(b) = sin(a) + sin(b)

We need to show,

+> cos(2a) + cos(2b) = 2cos($\pi$ + a + b) ----------  I

+> sin(2a) + sin(2b) = 2sin($\pi$ + a + b)    ----------  II

Now, we know, cos(2A) = $cos^{2}$(A) - $sin^{2}$(A),

Similarly, sin(2A) = 2 sin(A)cos(A),

Now, in I,

$cos^{2}$(a) - $sin^{2}$(a) +

$[cos(a) + cos(b)]^{2}$ - 2cos(a)cos(b) -  $sin^{2}$(a) -

From the given condition, $sin^{2}$(a) +

2sin(a)sin(b) - 2cos(a)cos(b)

2(-(cos(a)cos(b) - sin(a)sin(b)))

2cos($\pi$ + a + b)

Hence we proved the first part.

Before seeing the answer, try solving it on you own.

Concept :- Get the terms to the given condition (i.e) cos(a) + cos(b) = sin(a) + sin(b).

Hope You have solved it on your own once before seeing the solution.

Now, we use formula and get,

2 sin(a)cos(a) + 2 sin(b)cos(b)

2 sin(a)cos(a) [ 1 + [sin(b)cos(b)] / sin(a)cos(a) ] ------------ IV

Now, we tweek the condition as follows,

cos(a) + cos(b) = sin(a) + sin(b)

Dividing by cos(b)sin(b)

cos(a) / sin(b)cos(b) + 1 / sin(b) = sin(a) / cos(b)sin(b) + 1 / cos(b)

[cos(a) - sin(a)] / sin(b)cos(b) = [cos(b) - sin(b)] / cos(b)sin(b)

cos(a) - cos(b) = sin(a) - sin(b) --------------- III

Now add this to the given condition, we get

sin(a) = cos(a)

Then , sin(b) = cos(b)

So, now IV is easy to solve.

2 sin(a)cos(a) [ 1 + [sin(b)cos(b)] / sin(a)cos(a) ]

2 $sin^{2}$(a) [ 1 + $sin^{2}$(b) / $sin^{2}$(a) ]

2 [ $sin^{2}$(a) + $sin^{2}$(b) ]

2 [ sin(a)cos(a) + sin(b)cos(b) ]

2 sin(a+ b)

Now. in second quadrant, sin is positive.

2sin($\pi$ + a + b)

Hence, both the conditions are proved.