Using addition and subtraction identity prove that \[\sin \left( {A + B} \right).\sin \left( {A - B} \right) = {\sin ^2}A - {\sin ^2}B\]
Answer
582.3k+ views
Hint: This is simply a problem related to sum and difference identities. We will start with the LHS of the question and with the help of identities will proceed towards the RHS. We will try to make the terms in terms of sin function because the RHS is in sin function only.
Formula used:
\[\sin \left( {A + B} \right) = \sin A.\cos B + \cos A.\sin B\]
\[\sin \left( {A - B} \right) = \sin A.\cos B - \cos A.\sin B\]
Complete step-by-step answer:
We can directly start with the LHS as,
\[\sin \left( {A + B} \right).\sin \left( {A - B} \right)\]
Now we will write the formulas expansion as,
\[ = \left( {\sin A.\cos B + \cos A.\sin B} \right)\left( {\sin A.\cos B - \cos A.\sin B} \right)\]
Now will multiply the terms in first bracket with those in the second bracket,
\[ = \sin A.\cos B\left( {\sin A.\cos B - \cos A.\sin B} \right) + \cos A.\sin B\left( {\sin A.\cos B - \cos A.\sin B} \right)\]
On multiplying we get,
\[ = {\sin ^2}A.{\cos ^2}B - \sin A.\cos B.\cos A.\sin B + \cos A.\sin B.\sin A.\cos B - {\cos ^2}A.{\sin ^2}B\]
On careful observation we come to the conclusion that the second and third term are opposites of each other. So on cancelling them we get,
\[ = {\sin ^2}A.{\cos ^2}B - {\cos ^2}A.{\sin ^2}B\]
Now we can observe half of the RHS in the answer above only need to remove the cos function. So as we know,
\[{\cos ^2}\theta = 1 - {\sin ^2}\theta \]
Applying this to equation above,
\[ = {\sin ^2}A\left( {1 - {{\sin }^2}B} \right) - \left( {1 - {{\sin }^2}A} \right){\sin ^2}B\]
On multiplying again,
\[ = {\sin ^2}A - {\sin ^2}A.{\sin ^2}B - {\sin ^2}B + {\sin ^2}A.{\sin ^2}B\]
Cancelling the opposite terms,
\[ = {\sin ^2}A - {\sin ^2}B\]
And this is ,
\[ = RHS\]
Hence proved.
Note: Note that this is a very basic problem as the question itself helps us to lead towards the problem. Only the thing is when we multiply the brackets we should be careful about the terms and the signs. One wrong sign can collapse the proof and just wastes the time. Along with sum and difference formulas we should be familiar with factorization and defactorization formulae.
Formula used:
\[\sin \left( {A + B} \right) = \sin A.\cos B + \cos A.\sin B\]
\[\sin \left( {A - B} \right) = \sin A.\cos B - \cos A.\sin B\]
Complete step-by-step answer:
We can directly start with the LHS as,
\[\sin \left( {A + B} \right).\sin \left( {A - B} \right)\]
Now we will write the formulas expansion as,
\[ = \left( {\sin A.\cos B + \cos A.\sin B} \right)\left( {\sin A.\cos B - \cos A.\sin B} \right)\]
Now will multiply the terms in first bracket with those in the second bracket,
\[ = \sin A.\cos B\left( {\sin A.\cos B - \cos A.\sin B} \right) + \cos A.\sin B\left( {\sin A.\cos B - \cos A.\sin B} \right)\]
On multiplying we get,
\[ = {\sin ^2}A.{\cos ^2}B - \sin A.\cos B.\cos A.\sin B + \cos A.\sin B.\sin A.\cos B - {\cos ^2}A.{\sin ^2}B\]
On careful observation we come to the conclusion that the second and third term are opposites of each other. So on cancelling them we get,
\[ = {\sin ^2}A.{\cos ^2}B - {\cos ^2}A.{\sin ^2}B\]
Now we can observe half of the RHS in the answer above only need to remove the cos function. So as we know,
\[{\cos ^2}\theta = 1 - {\sin ^2}\theta \]
Applying this to equation above,
\[ = {\sin ^2}A\left( {1 - {{\sin }^2}B} \right) - \left( {1 - {{\sin }^2}A} \right){\sin ^2}B\]
On multiplying again,
\[ = {\sin ^2}A - {\sin ^2}A.{\sin ^2}B - {\sin ^2}B + {\sin ^2}A.{\sin ^2}B\]
Cancelling the opposite terms,
\[ = {\sin ^2}A - {\sin ^2}B\]
And this is ,
\[ = RHS\]
Hence proved.
Note: Note that this is a very basic problem as the question itself helps us to lead towards the problem. Only the thing is when we multiply the brackets we should be careful about the terms and the signs. One wrong sign can collapse the proof and just wastes the time. Along with sum and difference formulas we should be familiar with factorization and defactorization formulae.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

