Trigonometric Identities Question 9
Question: The expression $ {{\cos }^{2}}(A-B)+{{\cos }^{2}}B-2\cos (A-B)\cos A\cos B $ is
Options:
A) Dependent on B
B) Dependent on A and B
C) Dependent on A
D) Independent of A and B
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{\cos }^{2}}(A-B)+{{\cos }^{2}}B-2,\cos ,(A-B),\cos A\cos B $ $ ={{\cos }^{2}}(A-B)+{{\cos }^{2}}B $ $ -\cos ,(A-B),{ \cos (A-B)+\cos (A+B) } $ $ ={{\cos }^{2}}B-\cos ,(A-B)\cos (A+B) $ $ ={{\cos }^{2}}B-({{\cos }^{2}}A-{{\sin }^{2}}B)=1-{{\cos }^{2}}A $ Hence it depends on A. Trick: Put two different values of A. Let $ A=90^{o}, $ then the value of expression will be $ {{\sin }^{2}}B+{{\cos }^{2}}B=1 $ Now put $ A=0^{o} $ , then the value of expression will be $ {{\cos }^{2}}B+{{\cos }^{2}}B-2{{\cos }^{2}}B=0 $ It means that the expression has different values for different A i.e. it depends on A. Now similarly for $ B=90^{o}, $ the value of expression will be $ {{\sin }^{2}}A+0-0 $ $ ={{\sin }^{2}}A $ and at $ B=0^{o} $ the value of expression will be $ {{\cos }^{2}}A+1-2{{\cos }^{2}}A={{\sin }^{2}}A $ . Hence, the expression has the same value for different values of B, so it does not depend on B.
BETA
