Question

(sin^3A+sin3A/sinA)+(cos^3A-cos3A/cosA)​

Answer 1

Answer:

3

Step-by-step explanation:

$\frac{sin^{3}A +sin3A}{sinA}+\frac{cos^{3}A-cos3A }{cosA}$

sin3A can be written as

sin(2A+A) = sin2AcosA+cos2AsinA     ∵sin(A+B)=sinAcosB+cosAsinB

cos3A can be written as

cos(2A+A)=cos2AcosA-sin2AsinA       ∵cos(A+B)=cosAcosB-sinAsinB

so given equation can be written as

$\frac{sin^{3}A+sin2AcosA+cos2AsinA }{sinA}+\frac{cos^{3} A-(cos2AcosA-sin2AsinA)}{cosA}$

=$\frac{sin^{3}A }{sinA}+\frac{sin2AcosA}{sinA} +\frac{cos2AsinA}{sinA} +\frac{cos^{3}A }{cosA}- \frac{cos2AcosA}{cosA} +\frac{sin2AsinA}{cosA}$

we know that sin2A=2sinAcosA so the above equation becomes

=$sin^{2}A+cos^{2}A+\frac{2sinAcosAcosA}{sinA}+cos2A-cos2A+\frac{2sinAcosAsinA}{cosA}$

= 1+ 2cos²A+2sin²A

= 1+2(sin²A+cos²A)

=3                                                                  ∵sin²A+cos²A=1

Answer 2

Answer:

Required value of the given term is 3.

Step-by-step explanation:

Given,

$\frac{ {sin}^{3}A + \sin(3A) }{ \sin(A) } + \frac{ {sin}^{3}A - \cos(3A) }{cosA} ......(1)$

We know,

$sin3A = \sin(2A + A) = \sin(2A) \cos(A) + \sin(2A) \sin(A)$

Again

$\cos(3A) = \cos(2A +A ) = \cos(2A) \cos(A) - \sin(2A) \sin(A)$

Now from (1),

$\frac{ {sin}^{3}A + \sin(2A) \cos(A) + \cos(2A) \sin(A) }{ \sin(A) } + \frac{ { \cos}^{3}A - \cos(2A) \cos(A) + \sin(2A) \sin(A) }{ \cos(A) }$

We are separating every term,

$= \frac{ { \sin}^{3}A }{ \sin(A) } + \frac{ \sin(2A) \cos(A) }{ \sin(A) } + \frac{ \cos(2A) \sin(A) }{ \sin(A) } + \frac{ {cos}^{3} A }{ \cos(A) } - \frac{ \cos(2A \cos(A) ) }{ \cot(A) } + \frac{ \sin(2A) \sin(A) }{ \cos(A) }$

$= {sin}^{2} A + {cos}^{2} A + \frac{2 \sin(A) \cos(A) \cos(A) }{ \sin(A) } \cos(2A) - \cos(2A) + \frac{2 \sin(A) \cos(A) \sin(A) }{ \cos(A) }$

$= 1 + 2 {cos}^{2} A + 2 {sin}^{2} A$

$= 1 + 2 = 3$

Here applied formulas,

$sin(x + y) = \sin(x) \cos(y) + \sin(y) \cos(x)$

$\cos(x + y) = \cos(x) \cos(y) - \sin(x) \sin(y)$

$\sin(2x) = 2 \sin(x) \cos(x)$

${sin}^{2} x + {cos}^{2} x = 1$