Question

Which of the following is true for \[c\cos \left( {A - \alpha } \right) + a\cos \left( {C + \alpha } \right)\] in \[\Delta ABC\]?
A. \[a\cos \alpha \]
B. \[b\cos \alpha \]
C. \[c\cos \alpha \]
D. \[2b\cos \alpha \]

Answer 1

Hint: First we will apply the sum and difference formula of cos. Then use the cosine law and sine law to simplify the given expression.

Formula used:
Sine Law:
\[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}\]
Cosine law:
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]
Trigonometry identity:
\[\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B\]
\[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\]

Complete step by step solution:
Given expression is
\[c\cos \left( {A - \alpha } \right) + a\cos \left( {C + \alpha } \right)\]
Applying sum and difference formula of cosine in the above expression
\[ = c\cos A\cos \alpha + c\sin A\sin \alpha + a\cos C\cos \alpha - a\sin C\sin \alpha \]
\[ = c\cos A\cos \alpha + a\cos C\cos \alpha + c\sin A\sin \alpha - a\sin C\sin \alpha \]
Take common \[\cos \alpha \] from first two terms and \[\sin \alpha \] last two terms.
\[ = \cos \alpha \left( {c\cos A + a\cos C} \right) + \sin \alpha \left( {c\sin A - a\sin C} \right)\] ….(i)
Form cosine law we get
\[{a^2} = {b^2} + {c^2} - 2bc\cos A \Rightarrow \cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C \Rightarrow \cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\]
Substitute \[\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\] and \[\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\] in (i)
\[ = \cos \alpha \left( {c \cdot \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} + a \cdot \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}} \right) + \sin \alpha \left( {c\sin A - a\sin C} \right)\]
\[ = \cos \alpha \left( {\dfrac{{{b^2} + {c^2} - {a^2}}}{{2b}} + \dfrac{{{a^2} + {b^2} - {c^2}}}{{2b}}} \right) + \sin \alpha \left( {c\sin A - a\sin C} \right)\]
\[ = \cos \alpha \left( {\dfrac{{{b^2} + {c^2} - {a^2} + {a^2} + {b^2} - {c^2}}}{{2b}}} \right) + \sin \alpha \left( {c\sin A - a\sin C} \right)\]
\[ = \cos \alpha \left( {\dfrac{{2{b^2}}}{{2b}}} \right) + \sin \alpha \left( {c\sin A - a\sin C} \right)\]
\[ = b\cos \alpha + \sin \alpha \left( {c\sin A - a\sin C} \right)\]
We know the sine law \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = k\left( {say} \right)\]
Then, \[\sin A = ka\], \[\sin C = kc\]
\[ = b\cos \alpha + \sin \alpha \left( {kac - kac} \right)\]
\[ = b\cos \alpha \]
Hence option B is the correct option.

Note: Some students often confuse with the sum and difference of cos. They often used \[\cos \left( {A - B} \right) = \cos A\cos B - \sin A\sin B\] and \[\cos \left( {A + B} \right) = \cos A\cos B + \sin A\sin B\] which are incorrect formula. The correct formula is \[\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B\] and \[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\].