Prove that: cos³A.cos3A +sin³A.sin3A = cos²2A | Trigonometric Identities of Multiple Angles | Sci-Pi (cos³acos3a+sin³asin3a = cos³2A)

Trigonometric Identities of Multiple Angles | Sci-Pi
This is a class 10 Question From Values of Trigonometric Ratios chapter of Unit Trigonometry (Multiples Angles).

Solution:

We know, 


cos3A = 4cos³A-cos3A
So,
cos³A.cos3A = cos³A (4cos³A-3cosA)

= 4cos⁶A - 3cos⁴A --- (i)



sin3A = 3sinA -4sin³A
So,
sin³A.sin3A = sin³A (3sinA -4sin³A)
= 3sin⁴A -4sin⁶A ----- (ii)

cos⁶A - sin⁶A = (cos²A)³ - (sin²A)³
= (cos²A-sin²A)(cos⁴A+sin⁴A+cos²Asin²A)
= (cos2A){(cos²A+sin²A)² -2cos²Asin²A +cos²Asin²A)
= (cos2A)(1 -cos²Asin²A) ---- (iii)

cos⁴A -sin⁴A = (cos²A+sin²A)(cos²A-sin²A)
= 1(cos2A)
= cos2A ---- (iv)

- 4cos²Asin²A (cos²A -sin²A) 

= -4cos⁴Asin²A +4cos²Asin⁴A

= 4cos²Asin⁴A -4cos⁴Asin²A

= 3cos²Asin⁴A +cos²Asin⁴A -3cos⁴Asin²A -cos⁴Asin²A

= 3cos²Asin⁴A -3cos⁴Asin²A +cos²Asin⁴A -cos⁴Asin²A

= 3cos²Asin²A(sin²A -cos²A) +cos²Asin²A(sin²A -cos²A)

[Taking - common]

= - 3cos²Asin²A(cos²A -sin²A) -cos²Asin²A(cos²A -sin²A)

= - 3cos²Asin²A(cos2A) -cos²Asin²A(cos2A) ---- (v)





Our question is:


cos³A.cos3A +sin³A.sin3A = cos³2A

Taking LHS

cos³A.cos3A +sin³A.sin3A 

[Put the values from (i) and (ii)]

= 4cos⁶A - 3cos⁴A + 3sin⁴A -4sin⁶A

= 4cos⁶A  -4sin⁶A - 3cos⁴A +3sin⁴A

= 4(cos⁶A -sin⁶A) -3(cos⁴A-sin⁴A)

[Put the value of (cos⁶A -sin⁶A) from (iii) ]

= 4(cos2A)(1 -cos²Asin²A) -3(cos⁴A-sin⁴A)

[Put the values of (cos⁴A-sin⁴A) from (iv) ]

= 4cos2A(1 -cos²Asin²A) -3(cos2A)

= 4cos2A -4cos2Acos²Asin²A -3cos2A

= 4cos2A - 3cos2A - 4cos2Acos²Asin²A

= cos2A - 4(cos²A-sin²A)cos²Asin²A

= cos2A - 4cos²Asin²A (cos²A -sin²A)

[Put value of - 4cos²Asin²A (cos²A -sin²A) from (v) ]

= cos2A - 3cos²Asin²A(cos2A) -cos²Asin²A(cos2A) 

= cos2A - cos²Asin²A(cos2A) - 3cos²Asin²A(cos2A)

= cos2A (1 -cos²Asin²A) - 3cos²Asin²A(cos2A)

[ Put cos⁶A -sin⁶A instead of cos2A - cos2A (1 -cos²Asin²A) from (iii) ]

= cos⁶A -sin⁶A - 3cos²Asin²A(cos2A)

= (cos²A)³ - (sin²A)³ - 3cosAsinA(cos²A -sin²A)

[We know, (a-b)³ = a³-b³-3ab(a-b) ]

= (cos²A - sin²A)³

= (cos2A)³

= cos³2A 

= RHS
#proved


This is a FAQ question that has very little solution present on the Internet. We hope that this detailed solution will help you solving this question.

Some other indentities used in this solution are:

a²-b² = (a+b)(a-b)

(a³-b³) = (a-b)(a²+ab+b²)

a²+b² = (a+b)² -2ab

cos²A -sin²A = cos2A






#FAQ
#SciPi