Question

prove the given Identity of multiple of trigonometry​

Answer 1

Given to prove that :-

sin5A = 16sin⁵A - 20sin³A + 5sinA

Formulae used :-

sin(x +y) = sinx cosy + siny cosx

sin3x = 3sinx - 4 sin³A

cos3x = 4 cos³x - 3cosx

sin2x = 2sinx cosx

cos2x = 1 - 2sin²x [in terms of sin]

cos²A = 1- sin²A

By using these formulae we shall prove it

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sin5A can be written as sin(3A+2A) in order to get in form of sin(A+ B) So, expanding the sin5A according to the formula

sin(3A + 2A) = sin3A cos2A + sin2A cos3A

= (3 sinA - 4sin³A) (1 -2 sin²A) + (2sinA cosA)(4cos³A- 3cosA)

Multiplying the terms

3sinA (1 - 2sin²A) - 4sin³A(1 -2sin²A) + 2sinAcosA (4cos³A - 3 cosA)

3sinA - 6sin³A - 4sin³A + 8 sin⁵A + 8sinA cos⁴A - 6 sinA cos²A

Converting all "cos" terms into" sin" terms In order to get the R.H. S

3sinA - 10sin³A + 8sin⁵A + 8sinA (1-sin²A)² - 6 sinA (1 - sin²A)

Here cos²A can be written in terms of sin are (1-cos²A) i.e from

sin²A + cos²A = 1

sin²A = 1-cos²A

Expanding (1-sin²A) by using (a-b)² formula = a²-2ab + b²

3sinA - 10sin³A + 8sin⁵A + 8sinA( 1 + sin⁴A - 2sin²A) -6sinA + 6sin³A

3sinA - 10sin³A + 8sin⁵A + 8sinA + 8sin⁵A - 16 sin³A - 6sinA + 6sin³A

Keeping all like powers together

8sin⁵A + 8sin⁵A - 10sin³A - 16sin³A + 6sin³A + 3sinA + 8sinA - 6 sinA

16sin⁵A - 20sin³A + 5sinA

Hence ,

sin(3A + 2A) = 16sin⁵A - 20sin³A + 5sinA

sin5A = 16sin⁵A - 20sin³A + 5sinA

Hence proved!

Note :- Instead of theta Here I replaced with letter "A"