Question: If $ x+y=z, $ then the value of $ {{\cos }^{2}}x+{{\cos }^{2}}y+{{\cos }^{2}}z $ is
Options:
A) $ 1+2\sin x\sin y\sin z $
B) $ 1-2\sin x\sin y\sin z $
C) $ 1+2\cos x\cos y\cos z $
D) $ 1-2\cos x\cos y\cos z $
E) None of the above
Show Answer
Answer:
Correct Answer: C
Solution:
- Given, $ x+y=z $
Now, $ {{\cos }^{2}}x+{{\cos }^{2}}y+{{\cos }^{2}}z=? $
$ =1+({{\cos }^{2}}x-{{\sin }^{2}}y)+{{\cos }^{2}}z $
$ =(\cos x-\sin y)(\cos x+\sin y) $
$ =[ \cos x-\cos ( \frac{\pi }{2}-y ) ][ \cos x+\cos ( \frac{\pi }{2}-y ) ] $
$ =[ -2\sin \frac{( x-\frac{\pi }{2}+y )}{2}\sin \frac{( x+\frac{\pi }{2}-y )}{2} ] $
$ [ 2\cos \frac{( x+\frac{\pi }{2}-y )}{2}\cos \frac{( x-\frac{\pi }{2}+y )}{2} ] $
$ =-\sin ( x+\frac{\pi }{2}-y )\sin ( x-\frac{\pi }{2}+y ) $
$ =\cos (x-y)\cos (x+y) $
$ =1+\cos (x+y)\cdot \cos (x-y)+{{\cos }^{2}}z $
$ =1+\cos z\cos (x-y)+{{\cos }^{2}}z $
$ =1+\cos z[\cos (x-y)+cos(x+y)] $
$ =1+\cos z $
$ [ 2\cos \frac{(x-y+x+y)}{2}\cdot \cos \frac{(x-y-x-y)}{2} ] $
$ =1+2\cos z\cdot \cos x\cdot \cos y $
$ =1+2\cos x\cdot \cos y\cdot cosz $
BETA
