How do you simplify $f (θ) = cot 4 θ - sin 2 θ - 2 cos 8 θ$ $f(theta)=cot4theta-sin2theta-2cos8theta$ to trigonometric functions of a unit $θ$ $theta$ ?

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Answer 1 · 2018-08-02T12:57:39

See answer in the explanation.

Use expansions forming real and imaginary parts in the

expansions

from ${(cos θ + i sin θ)}^{n} = cos n θ + i sin n θ, n = 2, 4 and 8$ $( cos theta + i sin theta )^n= cos ntheta + i sin ntheta, n = 2, 4 and 8$ .

For brevity, $c = cos θ and s = sin θ$ $c = cos theta and s = sin theta$ .

$f = f_{1} + f_{2} + f_{3}$ $f = f_1 + f_2 + f_3$ ,

$f_{1} = cot 4 θ = \frac{cos (4 θ)}{sin (4 θ)}$ $f_1 = cot 4theta = cos (4theta)/sin (4theta)$

$= \frac{c^{4} - 6 c s (c^{2} - s^{2}) + s^{4}}{4 c s (c^{2} - s^{2})}$ $= ( c^4-6 c s ( c^2 - s^2 ) + s^4 )/(4 c s ( c^2 - s^2 ))$

$f_{2} = - sin 2 θ = - 2 c s$ $f_2 = - sin 2theta = - 2 c s$

$f_{3} = - 2 cos 8 θ$ $f_3 = - 2 cos 8theta$

$= - 2 (c^{8} - 28 c s (c^{6} - s^{6}) + 70 c^{2} s^{2} (c^{2} - s^{2}) + s^{8})$ $= - 2 (c^8 - 28 c s (c^6 - s^6 ) + 70 c^2 s^2 ( c^2 - s^2) +s^8 )$

$f = f_{1} + f_{2} + f_{3}$ $f = f_1 + f_2 + f_3$ .

Of course, simplification is possible, using

${(c^{2} + s^{2})}^{m} = 1, m = 1, 2, 3, 4 .$ $(c^2 + s^2 )^m = 1, m = 1, 2, 3, 4.$

How do you simplify f(θ)=cot4θ−sin2θ−2cos8θf(theta)=cot4theta-sin2theta-2cos8theta to trigonometric functions of a unit θtheta?