How do you simplify $sin 8 θ - tan 2 θ$ $sin8theta-tan2theta$ to trigonometric functions of a unit $θ$ $theta$ ?

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How do you simplify $sin 8 θ - tan 2 θ$ $sin8theta-tan2theta$ to trigonometric functions of a unit $θ$ $theta$ ?

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Answer 1 · 2018-07-30T03:51:32

$8 sin θ cos θ ({cos}^{2} θ - {sin}^{2} θ)$ $8 sin theta cos theta ( cos^2theta- sin^2theta )$
$(1 - 8 {cos}^{2} θ {sin}^{2} θ)$ $( 1 - 8 cos^2theta sin^2theta )$
$- \frac{2 tan θ}{1 - {tan}^{2} θ}$ $- ( 2 tan theta )/( 1 - tan^2theta )$

Explanation:

Using De Moivre and Binomial theorems,

equate imaginary parts in

$(cos 8 θ + i sin 8 θ) = {(cos θ + i sin θ)}^{8}$ $( cos 8theta + i sin 8theta ) = ( cos theta + i sin theta )^8$

$= (r e a l) C_{8} + i (r e a l) S_{8}$ $= ( real ) C_8 + i ( real ) S_8$ and get

$sin 8 θ = S_{8}$ $sin 8theta = S_8$

$= 8 C_{1} ({cos}^{7} θ sin θ - cos θ {sin}^{7} θ)$ $= 8C_1 ( cos^7 theta sin theta - cos theta sin^7theta )$

$- 8 C_{3} ({cos}^{5} θ {sin}^{3} θ - {cos}^{3} θ {sin}^{5} θ)$ $- 8C_3 ( cos^5theta sin^3theta - cos^3theta sin^5theta )$

$= 8 sin θ cos θ ({cos}^{6} θ - {sin}^{6} θ$ $= 8 sin theta cos theta ( cos^6theta - sin^6theta$

$- 7 {cos}^{2} θ {sin}^{2} θ ({cos}^{2} θ - {sin}^{2} θ))$ $-7cos^2theta sin^2theta ( cos^2theta - sin^2theta ))$ .

$= 8 sin θ cos θ ({cos}^{2} θ - {sin}^{2} θ)$ $= 8 sin theta cos theta ( cos^2theta- sin^2theta )$

$(1 - 8 {cos}^{2} θ {sin}^{2} θ)$ $( 1 - 8 cos^2theta sin^2theta )$

I have used $n C_{r} = n C_{n - r}, {cos}^{2} θ + {sin}^{2} θ = 1$ $nC_r = nC_(n-r), cos^2theta + sin^2theta = 1$ and

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3 - b^3 = ( a - b )(a^2 +ab + b^2 )$ , for this simplification.

Also $tan 2 θ = \frac{2 tan θ}{1 - {tan}^{2} θ}$ $tan 2theta = ( 2 tan theta )/( 1 - tan^2theta )$ .

And so,

$sin 8 θ - tan 2 θ$ $sin 8theta - tan 2theta$

$= 8 sin θ cos θ ({cos}^{2} θ - {sin}^{2} θ)$ $= 8 sin theta cos theta ( cos^2theta- sin^2theta )$

$(1 - 8 {cos}^{2} θ {sin}^{2} θ)$ $( 1 - 8 cos^2theta sin^2theta )$

$- \frac{2 tan θ}{1 - {tan}^{2} θ}$ $- ( 2 tan theta )/( 1 - tan^2theta )$

Verification is done, with $θ = \frac{π}{8}$ $theta = pi/8$ .

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How do you simplify $sin 8 θ - tan 2 θ$ $sin8theta-tan2theta$ to trigonometric functions of a unit $θ$ $theta$ ?

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