Question

If \[A = B = 45^\circ \], Show that \[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\].

Answer 1

Hint: In this question, we need to prove that \[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\] with the given condition that if \[A = B = 45^\circ \] .To prove this, we will first consider and solve the left-hand side of the given expression using the given condition. After that we will solve the right-hand side of the given expression using the condition. And hence we will get the required result.

Complete step-by-step answer:
The given condition is, If \[A = B = 45^\circ \]
That is \[A = 45^\circ \] and \[B = 45^\circ \]
And we have to prove that,
\[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B{\text{ }} - - - \left( 1 \right)\]
Now, first we will consider the left-hand side of the equation \[\left( 1 \right)\]
That is, \[\cos \left( {A + B} \right)\]
By substituting the value of \[A\] and \[B\] ,we get \[\cos \left( {45^\circ + 45^\circ } \right)\]
On adding, we get \[\cos \left( {90^\circ } \right)\]
Now, we know that the value of \[\cos \left( {90^\circ } \right)\] is \[0\]
Thus, we get \[\cos \left( {A + B} \right) = 0{\text{ }} - - - \left( 2 \right)\]
Now, we will consider the right-hand side of the equation \[\left( 1 \right)\]
That is, \[\cos A\cos B - \sin A\sin B\]
So, by substituting the value of \[A\] and \[B\] ,we get
\[\cos \left( {45^\circ } \right)\cos \left( {45} \right) - \sin \left( {45^\circ } \right)\sin \left( {45^\circ } \right)\]
Now, we know that the value of \[\cos \left( {45^\circ } \right)\] and \[\sin \left( {45^\circ } \right)\] is \[\dfrac{1}{{\sqrt 2 }}\]
By substituting the values, we get
\[\left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right) - \left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( {\dfrac{1}{{\sqrt 2 }}} \right)\]
On multiplying the terms, we get
\[\left( {\dfrac{1}{2}} \right) - \left( {\dfrac{1}{2}} \right)\]
On subtracting, we get
\[ = 0\]
Thus, we get \[\cos A\cos B - \sin A\sin B = 0{\text{ }} - - - \left( 3 \right)\]
Now, on comparing equation \[\left( 1 \right)\] and equation \[\left( 2 \right)\] we get,
\[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\]
\[ \Rightarrow 0 = 0\]
Therefore, left-hand side is equals to right-hand side
Hence, we have proved that
\[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\]

Note: The concept used to prove the given problem is trigonometric identities and ratios. Trigonometric identities are nothing, but they involve trigonometric functions including variables and constants. Few things should be kept in mind when we come across such questions. Firstly, solve the left-hand side and right-hand side separately to get the answer easily. And secondly the concepts of trigonometric functions and the general values of trigonometric functions should be well known which are mostly used. The common technique used in this problem is the substitution method with the use of trigonometric functions.