Question

prove that sin4π/8+sin43π/8+sin45π/8+sin47π/8=3/2​

Answer 1

EXPLANATION.

$\sf \implies sin^{4} \dfrac{\pi}{8} + sin^{4}\dfrac{3 \pi}{8} + sin^{4} \dfrac{5 \pi}{8} + sin^{4} \dfrac{7 \pi}{8} = \dfrac{3}{2}$

As we know that,

We can write equation as,

⇒ sin⁴5π/8 = sin(π - 3π/8) = sin⁴3π/8

⇒ sin⁴7π/8 = sin(π - π/8) = sin⁴π/8.

$\sf \implies sin^{4} \dfrac{\pi}{8} + sin^{4} \dfrac{3 \pi}{8} + sin^{4} \dfrac{3 \pi}{8} + sin^{4} \dfrac{\pi}{8}$

$\sf \implies 2sin^{4} \dfrac{\pi}{8} + 2sin^{4} \dfrac{3 \pi}{8}$

$\sf \implies 2 \bigg[sin^{4} \dfrac{\pi}{8} + sin^{4} \dfrac{3 \pi}{8} \bigg]$

$\sf \implies 2 \bigg[\bigg(sin^{2} \dfrac{\pi}{8} \bigg)^{2} + \bigg(sin^{2} \dfrac{3 \pi}{8} \bigg)^{2} \bigg]$

As we know that,

Formula of Sin²∅ = 1 - cos2∅/2.

Apply this formula in equation, we get.

$\sf \implies 2 \bigg[\bigg(\dfrac{1 - cos 2(\dfrac{\pi}{8} )}{2} \bigg)^{2} + \bigg(\dfrac{1 - cos 2 ( \dfrac{3 \pi}{8} )}{2} \bigg)^{2} \bigg]$

$\sf \implies 2 \bigg[\dfrac{(1 - Cos \dfrac{\pi}{4} )^{2} }{4} + \dfrac{ (1 - cos \dfrac{3 \pi}{4} )^{2} }{4} \bigg]$

$\sf \implies \dfrac{2}{4} \bigg[\bigg(1 - \dfrac{1}{\sqrt{2} } \bigg)^{2} + \bigg(1 + \dfrac{1}{\sqrt{2} } \bigg)^{2} \bigg]$

$\sf \implies \dfrac{1}{2} \bigg[ 1 + \dfrac{1}{2} - \dfrac{2}{\sqrt{2} } + 1 + \dfrac{1}{\sqrt{2} } + \dfrac{2}{\sqrt{2} } \bigg]$

$\sf \implies \dfrac{1}{2} \bigg[2 + 1 \bigg]$

$\sf \implies \dfrac{3}{2}$

                                                                                                                         

MORE INFORMATION.

(1) = sin2∅ = 2sin∅.cos∅ = 2tan∅/1 + tan²∅.

(2) = cos2∅ = cos²∅ - sin²∅ = 2cos²∅ - 1 = 1 - 2sin²∅ = 1 - tan²∅/1 + tan²∅.

(3) = tan2∅ = 2tan∅/1 - tan²∅.

(4) = sin3∅ = 3sin∅ - 4sin³∅.

(5) = cos3∅ = 4cos³∅ - 3cos∅.

(6) = tan3∅ = 3tan∅ - tan³∅/1 - 3tan²∅.