How do you simplify $3 cos 8 θ - 6 sin 4 θ$ $3cos8theta-6sin4theta$ to trigonometric functions of a unit $θ$ $theta$ ?

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How do you simplify $3 cos 8 θ - 6 sin 4 θ$ $3cos8theta-6sin4theta$ to trigonometric functions of a unit $θ$ $theta$ ?

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Answer 1 · 2016-11-25T03:34:36

$= 3 (2 {(2 {(2 {cos}^{2} θ - 1)}^{2} - 1)}^{2} - 1) - 12 sin θ cos θ (2 {cos}^{2} θ - 1)$ $=3(2(2(2cos^2theta-1)^2-1)^2-1)-12sin theta cos theta (2 cos^2 theta-1)$

Explanation:

$3 cos 8 θ - 6 sin 4 θ$ $3 cos 8theta- 6 sin 4theta$

$= 3 (2 {cos}^{2} 4 θ - 1) - 6 sin 2 θ cos 2 θ$ $=3(2 cos^2 4theta-1)-6 sin 2theta cos 2theta$

$= 3 (2 {(2 {cos}^{2} 2 θ - 1)}^{2} - 1) - 12 sin θ cos θ (2 {cos}^{2} θ - 1)$ $=3(2(2 cos^2 2theta-1)^2-1)-12 sin theta cos theta (2cos^2theta-1)$

$= 3 (2 {(2 {(2 {cos}^{2} θ - 1)}^{2} - 1)}^{2} - 1) - 12 sin θ cos θ (2 {cos}^{2} θ - 1)$ $=3(2(2(2cos^2theta-1)^2-1)^2-1)-12sin theta cos theta (2 cos^2 theta-1)$

Also, Comparison of real and imaginary parts in

$cos n θ + i sin n θ = {(cos θ + i sin θ)}^{n}, n = 4 and 8$ $cos ntheta+i sin ntheta = (cos theta+i sin theta)^n, n = 4 and 8$ can

be used as an alternative method. This is good, when m and n are

large, for expressions of the form $a cos m θ + b sin n θ)$ $a cos mtheta + b sin ntheta)$ .

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How do you simplify $3 cos 8 θ - 6 sin 4 θ$ $3cos8theta-6sin4theta$ to trigonometric functions of a unit $θ$ $theta$ ?

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How do you simplify 3cos8θ−6sin4θ3cos8theta-6sin4theta to trigonometric functions of a unit θtheta?

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How do you simplify $3 cos 8 θ - 6 sin 4 θ$ $3cos8theta-6sin4theta$ to trigonometric functions of a unit $θ$ $theta$ ?