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4 years ago

Derive identity sin(3x) = 3 sinx−4 sin³x

1 answer:

5 0

$sin3x=3sinx-4sin^3x\\\\L=sin(2x+x)=sin2xcosx+sinxcos2x\\\\=2sinxcosxcosx+sinx(cos^2x-sin^2x)\\\\=2sinxcos^2x+sinxcos^2x-sin^3x\\\\=3sinxcos^2x-sin^3x\\\\=3sinx(1-sin^2x)-sin^3x\\\\=3sinx-3sin^3x-sin^3x\\\\=3sinx-4sin^3x=R$

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Mary and Mia are assembling floor lamps. Mary can assemble 10 lamps per hour. Mia can complete 12 lamps per hour. Together, they

NNADVOKAT [17]

Answer:

Step-by-step explanation:

5 0

3 years ago

WORKSHEET-2

Margarita [4]

Answer:

1. $\frac{C}{D}$ = 3.143

2. Circumference

3. Area of the circle is 3850 $cm^{2}$ .

ii. Circumference of the circle is 220 cm.

Step-by-step explanation:

The area, A, and the circumference, c, of a circle can be determined respectively by:

A = $\pi$ $r^{2}$

C = 2 $\pi$ r

where r is the radius of the circle.

1. the ratio of the circumference and diameter, D, of a circle is:

r = $\frac{D}{2}$

so that,

C = 2 $\pi$ $\frac{D}{2}$

C = $\pi$ D

$\frac{C}{D}$ = $\pi$

= $\frac{22}{7}$

= 3.143

2. The distance around a circular region is known as it's circumference.

3. Given a circle of radius 35 cm, then:

A = $\pi$ $r^{2}$

= $\frac{22}{7}$ x $(35)^{2}$

= $\frac{22}{7}$ x 1225

= 22 x 175

A = 3850

Area of the circle is 3850 $cm^{2}$ .

C = 2 $\pi$ r

= 2 x $\frac{22}{7}$ x 35

= 2 x 22 x 5

= 220

C = 220 cm

Circumference of the circle is 220 cm.

8 0

3 years ago

Find the 7th term in the sequence?<br> 1/3,1,3,9

schepotkina [342]

Answer:

27 64 213 1004

Step-by-step explanation:

its 3 by 3 so its 213

3 0

3 years ago

What is 369+12354 pls

Anastaziya [24]

Answer:

The asnwer is 12723

Hope this helped!!!XD

7 0

3 years ago

Find the area of the surface generated by revolving the curve xequals=StartFraction e Superscript y Baseline plus e Superscript

artcher [175]

Solution :

$x=f(y) = \frac{e^y + e^{-y}}{2} , \ \ \ \ \ 0 \leq y \leq \ln 2$

$\frac{dx}{dy} = \frac{e^y + e^{-y}}{2}$

$\left(\frac{dx}{dy}\right)^2 = \frac{e^{2y} - 2 + e^{-2y}}{4}$

$1+\left(\frac{dx}{dy}\right)^2 = 1+\frac{e^{2y} - 2 + e^{-2y}}{4} = \frac{e^{2y} + 2 + e^{-2y}}{4}$

$= \left(\frac{e^y + e^{-y}}{2}\right)^2$

$\sqrt{1+\left(\frac{dx}{dy}\right)^2} = \sqrt{\left(\frac{e^y + e^{-y}}{2}\right)^2}=\frac{e^y + e^{-y}}{2}$

$S = \int_{y=a}^b 2 \pix \sqrt{1+\left(\frac{dx}{dy}\right)^2 } \ dy$

$=\int_{0}^{\ln2} 2 \pi \left(\frac{e^y+e^{-y}}{2}\right) \left(\frac{e^y+e^{-y}}{2}\right) \ dy$

$=\frac{\pi}{2}\int_{0}^{\ln 2}(e^y+e^{-y})^2 \ dy = \frac{\pi}{2}\int_{0}^{\ln 2}(e^{2y}+e^{-2y}+2) \ dy$

$=\frac{\pi}{2} \left[ \frac{e^{2y}}{2} + \frac{e^{-2y}}{-2} + 2y \right]_2^{\ln 2}$

$=\frac{\pi}{2} \left[ \left(\frac{e^{2 \ln 2}}{2} + \frac{e^{-2\ln2}}{-2} + 2 \ln2 \right) - \left( \frac{e^0}{2} + \frac{e^0}{-2}+0\right) \right]$

$=\frac{\pi}{2}\left[ \frac{e^{\ln4}}{2} - \frac{e^{\ln(1/4)}}{2} + \ln 4 - \left( \frac{1}{2} - \frac{1}{2} + 0 \right) \right]$

$=\frac{\pi}{2} \left[\frac{4}{2} -\frac{1/4}{2} + \ln 4 \right]$

$=\frac{\pi}{2} \left[ 2-\frac{1}{8} + \ln 4 \right]$

$=\left( \frac{15}{8} + \ln 4 \right) \frac{\pi}{2}$

Therefore, $S = \frac{15}{16} \pi + \pi \ln 2$

3 0

3 years ago