| If \[x+y=z,\]then the value of \[{{\cos }^{2}}x+{{\cos }^{2}}y+{{\cos }^{2}}z\]is |
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
Correct Answer: C
Solution :
| Given, \[x+y=z\] |
| Now, \[{{\cos }^{2}}x+{{\cos }^{2}}y+{{\cos }^{2}}z=?\] |
| \[=1+\,\,({{\cos }^{2}}x-{{\sin }^{2}}y)+{{\cos }^{2}}z\] |
| \[=(\cos x-\sin y)(\cos x+\sin y)\] |
| \[=\left[ \cos x-\cos \left( \frac{\pi }{2}-y \right) \right]\left[ \cos x+\cos \left( \frac{\pi }{2}-y \right) \right]\] |
| \[=\left[ -\,\,2\sin \frac{\left( x-\frac{\pi }{2}+y \right)}{2}\sin \frac{\left( x+\frac{\pi }{2}-y \right)}{2} \right]\] |
| \[\left[ 2\cos \frac{\left( x+\frac{\pi }{2}-y \right)}{2}\cos \frac{\left( x-\frac{\pi }{2}+y \right)}{2} \right]\] |
| \[=-\,\,\sin \left( x+\frac{\pi }{2}-y \right)\sin \left( x-\frac{\pi }{2}+y \right)\] |
| \[=\cos (x-y)\cos (x+y)\] |
| \[=1+\cos \,\,(x+y)\cdot \cos \,\,(x-y)+{{\cos }^{2}}z\] |
| \[=1+\cos z\cos \,\,(x-y)+{{\cos }^{2}}z\] |
| \[=1+\cos z\,\,[\cos \,\,(x-y)+cos\,\,(x+y)]\] |
| \[=1+\cos z\] |
| \[\left[ 2\cos \frac{(x-y+x+y)}{2}\cdot \cos \frac{(x-y-x-y)}{2} \right]\] |
| \[=1+2\cos z\cdot \cos x\cdot \cos y\] |
| \[=1+2\cos x\cdot \cos \,\,y\cdot cos\,\,z\] |
You need to login to perform this action.
You will be redirected in
3 sec
