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Question: How do you prove \[\cosh \left( A+B \right)=\cosh A\cdot \cosh B+\sinh B\cdot \sinh A\]?...

How do you prove cosh⁡(A+B)=cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A\cosh \left( A+B \right)=\cosh A\cdot \cosh B+\sinh B\cdot \sinh A?

Explanation

Solution

This question is from the topic of trigonometry. In this question, we will first understand the values of cosh⁡x\cosh x and sinh⁡x\sinh x. After that, we will use those values to the right side of the equation that we have to prove. After that, we will solve further questions and try to make the right side of the equation equal to the left side of the equation.

Complete step-by-step solution:
Let us solve this question.
In this question, we have asked to prove the equation
cosh⁡(A+B)=cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A\cosh \left( A+B \right)=\cosh A\cdot \cosh B+\sinh B\cdot \sinh A
Let us first know the value or we can say the formula of cosh⁡x\cosh x and sinh⁡x\sinh x.
The formula for cosh⁡x\cosh x is
cosh⁡x=ex+e−x2\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}
And the formula for sinh⁡x\sinh x is
sinh⁡x=ex−e−x2\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}
So, let us first start proving from the right side of the equation cosh⁡(A+B)=cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A\cosh \left( A+B \right)=\cosh A\cdot \cosh B+\sinh B\cdot \sinh A, we can write
cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A\cosh A\cdot \cosh B+\sinh B\cdot \sinh A
Now, let us put the values using the formulas of sinh⁡x\sinh x and cosh⁡x\cosh x that we have seen above. We can write the above term as
cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A=(eA+e−A2)⋅(eB+e−B2)+(eB−e−B2)⋅(eA−e−A2)\cosh A\cdot \cosh B+\sinh B\cdot \sinh A=\left( \dfrac{{{e}^{A}}+{{e}^{-A}}}{2} \right)\cdot \left( \dfrac{{{e}^{B}}+{{e}^{-B}}}{2} \right)+\left( \dfrac{{{e}^{B}}-{{e}^{-B}}}{2} \right)\cdot \left( \dfrac{{{e}^{A}}-{{e}^{-A}}}{2} \right)
We can write the above equation as
⇒cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A=12(eA+e−A)⋅12(eB+e−B)+12(eB−e−B)⋅12(eA−e−A)\Rightarrow \cosh A\cdot \cosh B+\sinh B\cdot \sinh A=\dfrac{1}{2}\left( {{e}^{A}}+{{e}^{-A}} \right)\cdot \dfrac{1}{2}\left( {{e}^{B}}+{{e}^{-B}} \right)+\dfrac{1}{2}\left( {{e}^{B}}-{{e}^{-B}} \right)\cdot \dfrac{1}{2}\left( {{e}^{A}}-{{e}^{-A}} \right)
We can write the above equation as
⇒cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A=14[(eA+e−A)⋅(eB+e−B)+(eB−e−B)⋅(eA−e−A)]\Rightarrow \cosh A\cdot \cosh B+\sinh B\cdot \sinh A=\dfrac{1}{4}\left[ \left( {{e}^{A}}+{{e}^{-A}} \right)\cdot \left( {{e}^{B}}+{{e}^{-B}} \right)+\left( {{e}^{B}}-{{e}^{-B}} \right)\cdot \left( {{e}^{A}}-{{e}^{-A}} \right) \right]
⇒cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A=14[(eA+e−A)(eB+e−B)+(eB−e−B)(eA−e−A)]\Rightarrow \cosh A\cdot \cosh B+\sinh B\cdot \sinh A=\dfrac{1}{4}\left[ \left( {{e}^{A}}+{{e}^{-A}} \right)\left( {{e}^{B}}+{{e}^{-B}} \right)+\left( {{e}^{B}}-{{e}^{-B}} \right)\left( {{e}^{A}}-{{e}^{-A}} \right) \right]
Now, we will use the foil method here. The foil method formula is: (a+b)(c+d)=ac+ad+bc+bd
So, we can write the above equation as
⇒cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A=14[(eAeB+eAe−B+e−AeB+e−Ae−B)+(eBeA−eBe−A−e−BeA+e−Be−A)]\Rightarrow \cosh A\cdot \cosh B+\sinh B\cdot \sinh A=\dfrac{1}{4}\left[ \left( {{e}^{A}}{{e}^{B}}+{{e}^{A}}{{e}^{-B}}+{{e}^{-A}}{{e}^{B}}+{{e}^{-A}}{{e}^{-B}} \right)+\left( {{e}^{B}}{{e}^{A}}-{{e}^{B}}{{e}^{-A}}-{{e}^{-B}}{{e}^{A}}+{{e}^{-B}}{{e}^{-A}} \right) \right]
The above equation can also be written as
⇒cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A=14[eAeB+eAe−B+e−AeB+e−Ae−B+eBeA−eBe−A−e−BeA+e−Be−A]\Rightarrow \cosh A\cdot \cosh B+\sinh B\cdot \sinh A=\dfrac{1}{4}\left[ {{e}^{A}}{{e}^{B}}+{{e}^{A}}{{e}^{-B}}+{{e}^{-A}}{{e}^{B}}+{{e}^{-A}}{{e}^{-B}}+{{e}^{B}}{{e}^{A}}-{{e}^{B}}{{e}^{-A}}-{{e}^{-B}}{{e}^{A}}+{{e}^{-B}}{{e}^{-A}} \right]
We know that if base are same and multiplied with each other, then powers can be added, for example: eAeB=eA+B{{e}^{A}}{{e}^{B}}={{e}^{A+B}}. Using this, we can write the above equation as
⇒cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A=14[eA+B+eA−B+e−A+B+e−A−B+eA+B−e−A+B−eA−B+e−A−B]\Rightarrow \cosh A\cdot \cosh B+\sinh B\cdot \sinh A=\dfrac{1}{4}\left[ {{e}^{A+B}}+{{e}^{A-B}}+{{e}^{-A+B}}+{{e}^{-A-B}}+{{e}^{A+B}}-{{e}^{-A+B}}-{{e}^{A-B}}+{{e}^{-A-B}} \right]
Now, we cancelled out the similar terms. We can write the above equation as
⇒cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A=14[eA+B+e−A−B+eA+B+e−A−B]\Rightarrow \cosh A\cdot \cosh B+\sinh B\cdot \sinh A=\dfrac{1}{4}\left[ {{e}^{A+B}}+{{e}^{-A-B}}+{{e}^{A+B}}+{{e}^{-A-B}} \right]
We can write the above equation as
⇒cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A=14[eA+B+eA+B+e−A−B+e−A−B]=14[2eA+B+2e−A−B]\Rightarrow \cosh A\cdot \cosh B+\sinh B\cdot \sinh A=\dfrac{1}{4}\left[ {{e}^{A+B}}+{{e}^{A+B}}+{{e}^{-A-B}}+{{e}^{-A-B}} \right]=\dfrac{1}{4}\left[ 2{{e}^{A+B}}+2{{e}^{-A-B}} \right]
The above equation can also be written as
⇒cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A=24[eA+B+e−A−B]=eA+B+e−A−B2\Rightarrow \cosh A\cdot \cosh B+\sinh B\cdot \sinh A=\dfrac{2}{4}\left[ {{e}^{A+B}}+{{e}^{-A-B}} \right]=\dfrac{{{e}^{A+B}}+{{e}^{-A-B}}}{2}
⇒cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A=eA+B+e−(A+B)2\Rightarrow \cosh A\cdot \cosh B+\sinh B\cdot \sinh A=\dfrac{{{e}^{A+B}}+{{e}^{-\left( A+B \right)}}}{2}
Now, using the formula: cosh⁡x=ex+e−x2\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}, we can write the above equation as
⇒cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A=cosh⁡(A+B)\Rightarrow \cosh A\cdot \cosh B+\sinh B\cdot \sinh A=\cosh \left( A+B \right)
Now, we can see that we have proved the equation cosh⁡(A+B)=cosh⁡A⋅cosh⁡B+sinh⁡B⋅sinh⁡A\cosh \left( A+B \right)=\cosh A\cdot \cosh B+\sinh B\cdot \sinh A.

Note: We should have a better knowledge in the topic of trigonometry to solve this type of question easily.
We should remember the following formulas:
sinh⁡x=ex−e−x2\sinh x=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}
cosh⁡x=ex+e−x2\cosh x=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}
We should know about the foil method. The formula for foil method is:
(a+b)(c+d)=ac+ad+bc+bd\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd