How do you simplify $f (θ) = sin 4 θ - cos 6 θ$ $f(theta)=sin4theta-cos6theta$ to trigonometric functions of a unit $θ$ $theta$ ?

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How do you simplify $f (θ) = sin 4 θ - cos 6 θ$ $f(theta)=sin4theta-cos6theta$ to trigonometric functions of a unit $θ$ $theta$ ?

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Answer 1 · 2018-03-15T12:58:26

${sin (θ)}^{6} - 15 {cos (θ)}^{2} {sin (θ)}^{4} - 4 cos (θ) {sin (θ)}^{3} + 15 {cos (θ)}^{4} {sin (θ)}^{2} + 4 {cos (θ)}^{3} sin (θ) - {cos (θ)}^{6}$ $sin(theta)^6-15cos(theta)^2sin(theta)^4-4cos(theta)sin(theta)^3+15cos(theta)^4sin(theta)^2+4cos(theta)^3sin(theta)-cos(theta)^6$

Explanation:

We'll use the following two identities:
$sin (A \pm B) = sin A cos B \pm cos A sin B$ $sin(A+-B)=sinAcosB+-cosAsinB$
$cos (A \pm B) = cos A cos B \mp sin A sin B$ $cos(A+-B)=cosAcosB∓sinAsinB$

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

$cos (6 θ) = {cos}^{2} (3 θ) - {sin}^{2} (3 θ)$ $cos(6theta)=cos^2(3theta)-sin^2(3theta)$
$= {(cos (2 θ) cos (θ) - sin (2 θ) sin (θ))}^{2} - {(sin (2 θ) cos (θ) + cos (2 θ) sin (θ))}^{2}$ $=(cos(2theta)cos(theta)-sin(2theta)sin(theta))^2-(sin(2theta)cos(theta)+cos(2theta)sin(theta))^2$
$= {(cos (θ) ({cos}^{2} (θ) - {sin}^{2} (θ)) - 2 {sin}^{2} (θ) cos (θ))}^{2} - (2 {cos}^{2} (θ) sin (θ) + sin (θ) {({cos}^{2} (θ) - {sin}^{2} (θ))}^{2}$ $=(cos(theta)(cos^2(theta)-sin^2(theta))-2sin^2(theta)cos(theta))^2-(2cos^2(theta)sin(theta)+sin(theta)(cos^2(theta)-sin^2(theta))^2$
$= {({cos}^{3} (θ) - {sin}^{2} (θ) cos (θ) - 2 {sin}^{2} (θ) cos (θ))}^{2} - {(2 {cos}^{2} (θ) sin (θ) + {cos}^{2} (θ) sin (θ) - {sin}^{3} (θ))}^{2}$ $=(cos^3(theta)-sin^2(theta)cos(theta)-2sin^2(theta)cos(theta))^2-(2cos^2(theta)sin(theta)+cos^2(theta)sin(theta)-sin^3(theta))^2$
$= {({cos}^{3} (θ) - 3 {sin}^{2} (θ) cos (θ))}^{2} - {(3 {cos}^{2} (θ) sin (θ) - {sin}^{3} (θ))}^{2}$ $=(cos^3(theta)-3sin^2(theta)cos(theta))^2-(3cos^2(theta)sin(theta)-sin^3(theta))^2$
$= {cos}^{6} (θ) - 6 {sin}^{2} (θ) {cos}^{4} (θ) + 9 {sin}^{4} (θ) {cos}^{2} (θ) - 9 {sin}^{2} (θ) {cos}^{4} (θ) + 6 {sin}^{4} (θ) {cos}^{2} (θ) - {sin}^{6} (θ)$ $=cos^6(theta)-6sin^2(theta)cos^4(theta)+9sin^4(theta)cos^2(theta)-9sin^2(theta)cos^4(theta)+6sin^4(theta)cos^2(theta)-sin^6(theta)$

$sin(4theta)-cos(6theta)=4sin(theta)cos^3(theta)-4sin^3(theta)cos(theta)-(cos^6(theta)-6sin^2(theta)cos^4(theta)+9sin^4(theta)cos^2(theta)-9sin^2(theta)cos^4(theta)+6sin^4(theta)cos^2(theta)-sin^6(theta))$
$=4sin(theta)cos^3(theta)-4sin^3(theta)cos(theta)-cos^6(theta)+6sin^2(theta)cos^4(theta)-9sin^4(theta)cos^2(theta)+9sin^2(theta)cos^4(theta)-6sin^4(theta)cos^2(theta)+sin^6(theta)$
$=sin(theta)^6-15cos(theta)^2sin(theta)^4-4cos(theta)sin(theta)^3+15cos(theta)^4sin(theta)^2+4cos(theta)^3sin(theta)-cos(theta)^6$

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How do you simplify $f (θ) = sin 4 θ - cos 6 θ$ $f(theta)=sin4theta-cos6theta$ to trigonometric functions of a unit $θ$ $theta$ ?

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