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1 year ago

If sin A + sin^3 A = cos^2 A, prove that cos^6 A - 4cos^4 A + 8cos^2 A = 4.

Mathematics

2 answers:

ZanzabumX [31]1 year ago

6 0

Answer:

Step-by-step explanation:

sinA+sin

A=cos

⇒sinA[1+sin

A]=cos

⇒sinA[1+1−cos

A]=cos

squaring on both sides.

⇒sin

A[4+cos

A−4cos

A]=cos

⇒(1−cos

A)[4+cos

A−9cos

A]=cos

⇒4+cos

A−4cos

A−64cot

A−cos

A+9cos

A=cos

⇒

cos

A−4cos

A+8cos

A=4

Hence Prove

Sphinxa [80]1 year ago

4 0

Answer:

Step-by-step explanation:

sinA(1+sin^2A) = cos^2A

sinA(2 -cos^2A) = cos^2A

Squaring both sides,

sin^2A(4-4cos^2A +cos^4A) = cos^4A

(1-cos^2A)(4-4cos^2A +cos^4A) = cos^4A

4-4cos^2A +cos^4A-4cos^2A+4cos^4A-cos^6A = cos^4A

4 -cos^6A +4cos^4A -8cos^2A = 0

cos^6A - 4 cos^4A + 8cos^2A = 4

hence proveproven

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Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$