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Question: Prove that \[\sin (C + D) \times \sin (C - D) = {\sin ^2}C - {\sin ^2}D\] ?...

Prove that sin⁡(C+D)×sin⁡(C−D)=sin⁡2C−sin⁡2D\sin (C + D) \times \sin (C - D) = {\sin ^2}C - {\sin ^2}D ?

Explanation

Solution

Hint : Here in this question we have to prove the given inequality which is given in this question. This question involves the trigonometric function we should know about the trigonometry ratios and addition formulas of trigonometry functions. Hence by using the simple calculations we are going to prove the given inequality.

Complete step-by-step answer :
In the trigonometry we have six trigonometry ratios namely sine , cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot. The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosine is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent.
The addition formulas of the trigonometric functions is given by
sin⁡(a+b)=sin⁡a.cos⁡b+cos⁡a.sin⁡b\sin (a + b) = \sin a.\cos b + \cos a.\sin b
sin⁡(a−b)=sin⁡a.cos⁡b−cos⁡a.sin⁡b\sin (a - b) = \sin a.\cos b - \cos a.\sin b
cos⁡(a+b)=cos⁡a.cos⁡b−sin⁡a.sin⁡b\cos (a + b) = \cos a.\cos b - \sin a.\sin b
cos⁡(a−b)=cos⁡a.cos⁡b+sin⁡a.sin⁡b\cos (a - b) = \cos a.\cos b + \sin a.\sin b
Now consider the given question
LHSLHS
⇒sin⁡(C+D)×sin⁡(C−D)\Rightarrow \sin (C + D) \times \sin (C - D)
Since the above inequality is in the form of addition formulas of the trigonometric functions, by using the formulas the above inequality is written as
⇒(sin⁡Ccos⁡D+cos⁡Csin⁡D)×(sin⁡Ccos⁡D−cos⁡Csin⁡D)\Rightarrow \left( {\sin C\cos D + \cos C\sin D} \right) \times \left( {\sin C\cos D - \cos C\sin D} \right)
The above inequality is in the form of the algebraic identity that is (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2} , on considering this algebraic identity the above inequality is written as
⇒(sin⁡Ccos⁡D)2−(cos⁡Csin⁡D)2\Rightarrow {(\sin C\cos D)^2} - {(\cos C\sin D)^2}
⇒sin⁡2Ccos⁡2D−cos⁡2Csin⁡2D\Rightarrow {\sin ^2}C{\cos ^2}D - {\cos ^2}C{\sin ^2}D
From the trigonometry identities we have sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1
In trigonometry identity taking sin⁡2A{\sin ^2}A to RHS and it is written as 1−sin⁡2A=cos⁡2A1 - {\sin ^2}A = {\cos ^2}A
Therefore we have
⇒sin⁡2C(1−sin⁡2D)−(1−sin⁡2C)sin⁡2D\Rightarrow {\sin ^2}C(1 - {\sin ^2}D) - (1 - {\sin ^2}C){\sin ^2}D
On multiplying the terms we have
⇒sin⁡2C−sin⁡2Csin⁡2D−(sin⁡2D−sin⁡2Dsin⁡2C)\Rightarrow {\sin ^2}C - {\sin ^2}C{\sin ^2}D - ({\sin ^2}D - {\sin ^2}D{\sin ^2}C)
On simplifying we have
⇒sin⁡2C−sin⁡2Csin⁡2D−sin⁡2D+sin⁡2Dsin⁡2C\Rightarrow {\sin ^2}C - {\sin ^2}C{\sin ^2}D - {\sin ^2}D + {\sin ^2}D{\sin ^2}C
The term sin⁡2Csin⁡2D{\sin ^2}C{\sin ^2}D will get cancels and therefore we have
⇒sin⁡2C−sin⁡2D\Rightarrow {\sin ^2}C - {\sin ^2}D
⇒RHS\Rightarrow RHS
Here we have proved LHS = RHS.

Note : The question involves the trigonometric functions and we have to prove the trigonometric function. When we simplify the trigonometric functions and which will be equal to the RHS then the function is proved. While simplifying the trigonometric functions we must know about the trigonometric ratios and the trigonometric identities.
sin⁡2A+cos⁡2A=1{\sin ^2}A + {\cos ^2}A = 1 this identity must be used accordingly to get the required answer asked, using it in different ways will lead to some other results which though correct will not bring required to prove.