Proving x in Sin(pi*cosx) = Cos(pi*sinx)

[vaishakh]

My doubt lies with the following question. It is given that sin(pi*cosx) = cos(pi*sinx).
From the following information it is to be proved that x = (1/4)[(2n+1)pi +or- cos^-1(1/8)] where n is any natural number.



From the question we can make it clear that the range of pi*cosx and pi*sinx is between pi and -pi. So I took two cases. In the first case bot are positive. When they are positive they must lie in [0,pi/2]. Thus from the equation we get that pi*cosx and pi*sinx must be complementary.
Therefore pi*cosx = pi/2 - pi*sinx. Therefore cosx = (1/2) - sinx. Therefore sinx + cosx = (1/2) and this is the maxima of the function thus bringing up that x is 2n*pi + pi/4 in case1 while it is 2n*pi -3pi/4 in case2.
So that is something that is against my question or infact I have disproved what I should prove. Can anyone help?

 

[BerryBoy]

Just want to clear something up:
At University, we are seriously discouraged from writing, Sin^-1 /
Cos^-1 / Tan^-1, because they can mean two things.

First it, in your question, it could equal arccos(1/8):
(The angle created from a right-angled triangle with adjacent length 1 & hypotenuse length 8).

Second it can equal sec(1/8):
(The inverse of the value cos(1/8) i.e. (Cos 1/8)^-1

Obviously this is a special case, and you can write Sin^2(x) without confusion. Could you clarify which cos the question is referring to?

Regards,
Sam

 

[VietDao29]

From the question we can make it clear that the range of pi*cosx and pi*sinx is between pi and -pi. So I took two cases. In the first case bot are positive. When they are positive they must lie in [0,pi/2]. Thus from the equation we get that pi*cosx and pi*sinx must be complementary.
Therefore pi*cosx = pi/2 - pi*sinx. Therefore cosx = (1/2) - sinx. Therefore sinx + cosx = (1/2) and this is the maxima of the function thus bringing up that x is 2n*pi + pi/4 in case1 while it is 2n*pi -3pi/4 in case2.

The bolded part is wrong!
Are you sure that:
\sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{2}??
In fact:
\sin \frac{\pi}{4} + \cos \frac{\pi}{4} = \frac{1}{\sqrt 2} + \frac{1}{\sqrt 2} = \sqrt 2
To solve for x in:
\sin x + \cos x = \frac{1}{2}, you can just use the identity:
\sin x + \cos x = \sqrt 2 \cos \left( x - \frac{\pi}{4} \right).
So:
\sin x + \cos x = \frac{1}{2}
\Leftrightarrow \sqrt 2 \cos \left( x - \frac{\pi}{4} \right) = \frac{1}{2}
\Leftrightarrow \cos \left( x - \frac{\pi}{4} \right) = \frac{1}{2 \sqrt 2}.
You should be able to go from here, right?

 

[vaishakh]

sinx + cosx = 2^(1/2)cos(x-pi/4)
is this the modification of any standard identity or something of university level. i cannot recall it from something i learnt.

 

[VietDao29]

A short proof:
\sin x + \cos x = \sqrt 2 \left( \frac{1}{\sqrt 2} \sin x + \frac{1}{\sqrt 2} \cos x \right) = \sqrt 2 \left( \sin \frac{\pi}{4} \sin x + \cos \frac{\pi}{4} \cos x \right) = \sqrt 2 \cos \left( x - \frac{\pi}{4} \right)
---------------
To solve trig equation in form : a sin x + b cos x = c. Where a, b, c are constants. We often divide boths sides by \sqrt{a ^ 2 + b ^ 2}
Then let:
\sin \alpha = \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \quad \mbox{and} \quad \cos \alpha = \frac{b}{\sqrt{a ^ 2 + b ^ 2}}
Or let:
\cos \alpha = \frac{a}{\sqrt{a ^ 2 + b ^ 2}} \quad \mbox{and} \quad \sin \alpha = \frac{b}{\sqrt{a ^ 2 + b ^ 2}}
So:
\sin x + \cos x = \frac{1}{2}
\Leftrightarrow \frac{1}{\sqrt 2} \sin x + \frac{1}{\sqrt 2} \cos x = \frac{1}{2 \sqrt 2}
\Leftrightarrow \sin \frac{\pi}{4} \sin x + \cos \frac{\pi}{4} \cos x = \frac{1}{2 \sqrt 2}
\Leftrightarrow \cos \left( x - \frac{\pi}{4} \right) = \frac{1}{2 \sqrt 2}