Question

Prove the following :
$cos\left( A – B \right) = cosA. cosB\ + \ sinA. sinB$ if $A = B = 60^{o}$

Answer 1

In this question, we need to prove that $cos\left( A – B \right) = cosA. cosB\ + \ sinA. sinB$ and also given a condition that if $A = B = 60^{o}$ . First we can consider and solve the left part of the given expression and then we can solve the right part of the expression using the given condition. By using the trigonometric identities and functions, we can easily prove the given expression.

Complete step by step answer:
To prove,
$cos\left( A – B \right) = cosA.cosB\ + \ sinA.sinB$
Condition : if $A = B = 60^o$
That is $A = 60^o$ and $B = 60^o$
First we can consider the left part,
⇒ $cos(A – B)\$
By substituting the values of $A$ and $B$,
We get,
⇒ $cos(60^o – 60^o)$
By subtracting,
We get,
⇒ $cos0^o$
We know that $cos0^{o} = 1$
Thus we get $cos(A – B)\ = 1$ ••• (1)
Now we can consider the right part of the expression.
⇒ $cosA. cosB\ + \ sinA. sinB$
By substituting the values of $A$ and $B$,
We get,
⇒ $\cos\left( 60^o \right) \times \cos\left( 60^o \right) + \sin\left( 60^o \right) \times sin(60^o)$
We know that,
$cos60^o = \dfrac{1}{2}$
$sin60^o = \dfrac{\sqrt{3}}{2}$
By substituting the values,
We get,
⇒ $\left( \dfrac{1}{2} \right)\left( \dfrac{1}{2} \right) + \left( \dfrac{\sqrt{3}}{2} \right)\left( \dfrac{\sqrt{3}}{2} \right)\$
By multiplying,
We get,
⇒ $\left( \dfrac{1}{4} \right) + \left( \dfrac{3}{4} \right)$
By adding ,
We get,
⇒ $\dfrac{4}{4}$
By simplifying,
We get,
⇒ $cosA. cosB\ + \ sinA.sinB = 1$ ••• (2)
By equating (1) and (2) ,
We get
$cos\left( A – B \right) = cosA.cosB\ + \ sinA.sinB$
⇒ $1 = 1$
Thus we have proved.
$cos\left( A – B \right) = cosA.cosB\ + \ sinA.sinB$

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. The common technique used in this problem is the substitution method with the use of trigonometric functions.