Respuesta :

msm555

Answer:

[tex]= \dfrac{1}{16} \left(5-7 \cos (2\theta) - cos(4 \theta)(3+ \cos (2\theta) ) \right)[/tex]

Step-by-step explanation:

We want to rewrite the expression  [tex]\sin^6{\theta}[/tex] as a sum or difference of first powers with multiple angles using power-reducing formulas.

Apply the power-reducing formula for [tex]\sin^2{\theta}[/tex]:

[tex]\sin^6{\theta} = (\sin^2{\theta})^3[/tex]

Using formula for [tex]\sin^2{\theta}[/tex] in terms of double angle:

[tex]\Large \boxed{\boxed{ \sin^2{\theta} = \frac{1 - \cos{2\theta}}{2} }}[/tex]

Substitute this value:

We get:

[tex]\begin{aligned} \sin^6 \theta &=\left(\dfrac{1-\cos(2 \theta )}{2}\right)^3 \\\\ &=\dfrac{1}{8}(1-\cos(2\theta ))^3 \end{aligned}[/tex]

Apply the power-reducing formula for [tex]\cos^2{\theta}[/tex]:

[tex]= \dfrac{1}{8}(1-\cos(2\theta))^3[/tex]

[tex]= \dfrac{1}{8}\left( 1^3 - 3\cdot 1^2 \cdot cos(2\theta) + 3 \cdot 1 \cdot cos(2\theta) ^2 - cos(2\theta) ^3\right)[/tex]

[tex]= \dfrac{1}{8}\left( 1 - 3 cos(2\theta) + 3cos^2 (2\theta) - cos^3 (2\theta) \right)[/tex]

[tex]= \dfrac{1}{8}\left( 1 - 3 cos(2\theta) + 3cos^2 (2\theta) -cos (2\theta) cos^2 (2\theta) \right)[/tex]

Using identity of  [tex]\boxed{\boxed{cos^2 ( \theta)= \left( \dfrac{1 + cos(2 \theta)}{2}\right)}}[/tex] , we get

[tex]= \dfrac{1}{8}\left( 1 - 3 cos(2\theta) + 3\cdot \left( \dfrac{1 + cos( 4 \theta)}{2}\right) -cos (2\theta) \left( \dfrac{1 + cos(4 \theta)}{2}\right) \right)[/tex]

[tex]= \dfrac{1}{8}\left( 1 - 3 cos(2\theta) + \left( \dfrac{3 + 3cos(4 \theta)}{2}\right) - \left( \dfrac{cos (2\theta) +cos (2\theta) \ cos(4 \theta)}{2}\right) \right)[/tex]

[tex]= \dfrac{1}{8}\cdot \dfrac{ ( 2(1 - 3 cos(2\theta) ) + (3 + 3cos(4 \theta) - \cos (2\theta) - \cos (2\theta) cos(4 \theta)}{2} \right)[/tex]

[tex]= \dfrac{1}{8\cdot 2 } \left( 2 - 6cos(2\theta) + 3 + 3cos(4 \theta) - \cos (2\theta) - \cos (2\theta) cos(4 \theta)\right)[/tex]

[tex]= \dfrac{1}{16} \left( 2 + 3 - 6 cos(2\theta) - \cos (2\theta) + 3cos(4 \theta) - \cos (2\theta) cos(4 \theta) \right)[/tex]

[tex]= \dfrac{1}{16} \left(5-7 \cos (2\theta) - cos(4 \theta)(3+ \cos (2\theta) ) \right)[/tex]

So, the expression as a sum or difference of first powers

with multiple angles is:

[tex]\Large\boxed{\boxed{ \dfrac{1}{16} \left(5-7 \cos (2\theta) - cos(4 \theta)(3+ \cos (2\theta) ) \right)}}[/tex]

semsee45

Answer:

[tex]\dfrac{5}{16}-\dfrac{7}{16}\cos2\theta+\dfrac{3}{16}\cos4\theta-\dfrac{1}{16}\cos2\theta\cos4\theta[/tex]

Step-by-step explanation:

The Power Reducing Formulas in trigonometry enable us to express the  square of a trigonometric function in terms of smaller powers.

[tex]\boxed{\begin{array}{c}\underline{\textsf{Power Reducing Formulas}}\\\\\sin^2\theta=\dfrac{1}{2}(1-\cos2\theta)\\\\\\\cos^2\theta=\dfrac{1}{2}(1+\cos2\theta)\\\\\\\tan^2\theta=\dfrac{(1-\cos2\theta)}{(1+\cos2\theta)}\end{array}}[/tex]

To express sin⁶θ in terms of first powers, begin by rewriting it as a product of squares:

[tex]\sin^6\theta= \sin^2\theta \cdot (\sin^2\theta)^2[/tex]

Substitute the sine-squared formula:

[tex]\dfrac{1}{2}(1-\cos2\theta) \cdot \left(\dfrac{1}{2}(1-\cos2\theta)\right)^2[/tex]

[tex]\dfrac{1}{2}(1-\cos2\theta) \cdot \dfrac{1}{4}(1-\cos2\theta)^2[/tex]

[tex]\dfrac{1}{8}(1-\cos2\theta)(1-\cos2\theta)^2[/tex]

Expand (1 - cos2θ)²:

[tex]\dfrac{1}{8}(1-\cos2\theta)(1-\cos2\theta)(1-\cos2\theta)\\\\\\\\\dfrac{1}{8}(1-\cos2\theta)(1-2\cos2\theta+\cos^22\theta)[/tex]

Use the cosine-squared formula:

[tex]\dfrac{1}{8}(1-\cos2\theta)\left(1-2\cos2\theta+\dfrac{1}{2}(1+\cos4\theta)\right)[/tex]

[tex]\dfrac{1}{8}(1-\cos2\theta)\left(\dfrac{3}{2}-2\cos2\theta+\dfrac{1}{2}\cos4\theta\right)[/tex]

Factor out 1/2 from the second set of parentheses:

[tex]\dfrac{1}{8}(1-\cos2\theta)\cdot \dfrac{1}{2}\left(3-4\cos2\theta+\cos4\theta\right)[/tex]

[tex]\dfrac{1}{16}(1-\cos2\theta)\left(3-4\cos2\theta+\cos4\theta\right)[/tex]

Expand:

[tex]\dfrac{1}{16}(3-4\cos2\theta+\cos4\theta-3\cos2\theta+4\cos^22\theta-\cos2\theta\cos4\theta)[/tex]

[tex]\dfrac{1}{16}(3-7\cos2\theta+\cos4\theta+4\cos^22\theta-\cos2\theta\cos4\theta)[/tex]

Again, use the cosine-squared formula:

[tex]\dfrac{1}{16}\left(3-7\cos2\theta+\cos4\theta+4\left(\dfrac{1}{2}(1+\cos4\theta)\right)-\cos2\theta\cos4\theta\right)[/tex]

[tex]\dfrac{1}{16}\left(3-7\cos2\theta+\cos4\theta+2+2\cos4\theta-\cos2\theta\cos4\theta\right)[/tex]

[tex]\dfrac{1}{16}\left(5-7\cos2\theta+3\cos4\theta-\cos2\theta\cos4\theta\right)[/tex]

[tex]\dfrac{5}{16}-\dfrac{7}{16}\cos2\theta+\dfrac{3}{16}\cos4\theta-\dfrac{1}{16}\cos2\theta\cos4\theta[/tex]

Therefore, sin⁶θ rewritten as a sum or difference of first powers with multiple angles is:

[tex]\large\boxed{\boxed{\dfrac{5}{16}-\dfrac{7}{16}\cos2\theta+\dfrac{3}{16}\cos4\theta-\dfrac{1}{16}\cos2\theta\cos4\theta}}[/tex]

Ver imagen semsee45