trig (1 Viewer)

ozidolroks · May 11, 2009

can someone please explain this question to me . i know how to do the first part, i just don't know how to do the hence section.

1. by expanding sin (A+B) show that sin3@ (theta) = 3sin@-4sin^3@.
Hence or otherwise find the smallest value of x that satisfies the equation

12x^3- 9x+2=0

answer in 2 dp.

2. Using sin(A+B)= sinAcosB+ cocsAsinB prove that:
sin^3x-1/4(3sinx-sin^3x)

thankx

rheyn · May 11, 2009

ozidolroks said:
1. by expanding sin (A+B) show that sin3@ (theta) = 3sin@-4sin^3@.

Hence or otherwise find the smallest value of x that satisfies the equation
12x^3- 9x+2=0
answer in 2 dp.

$sin(\Theta + 2\Theta ) = sin\Theta cos2\Theta +cos\Theta sin2\Theta = sin\Theta (1-2sin^2\Theta ) + cos\Theta 2sin\Theta cos\Theta =sin\Theta -2sin^3\Theta +2sin\Theta (1-sin^2\Theta ) = sin\Theta -2sin^3\Theta +2sin\Theta -2sin^3\Theta = 3sin\Theta -4sin^3\Theta$

$12x^3-9x+2=0$

let x=sin (theta)

$\therefore 12sin^3\Theta -9sin\Theta +2=0$
$-3(4sin^3\Theta -3sin\Theta )=-2$
$sin3\Theta = 2/3$
$\therefore 3\Theta =41.81$
$\therefore \Theta =13.94^0$
$\therefore x= sin (13.94)=0.24$

For question 2, What do you want to prove? All you've given is an expression, not an equation.

Drongoski · May 11, 2009

If no one wants to do this I'll try (1)

$\100dpi Let\ \ sin3\theta \ = \ 3sin\theta - 4sin^{3}\theta \ =\ \frac{2}{3} \ \ \cdots \ \ [1]\\ \\ Solving\ \ sin3\theta \ =\ \frac{2}{3}\ \ u\ get\ smallest\ 3\theta\,\ div\ by\ 3\ get\ sin\ of\ this\ = 0.24085 .. \\ \\ \ x=0.24\ \ (2 d.p.)\ is\ the\ reqd\ soln.\ Why?\ \\ \\ By [1]:\ \ \frac{2}{3}\ =\ 3sin\theta \ - 4sin^{3}\theta \\ \\ \therefore \ \ 12 sin^{3}\theta \ - \ 9sin\theta\ +\ 2\ \ =\ \ 0\\ \\ This\ eqn\ is\ :\ \ 12x^{3}\ -\ 9x\ +\ 2\ =\ 0\ \ \ if\ we\ subst\ x=sin\theta \\ \\ Hence\ soln\ \ x\ =\ 0.24$

tommykins · May 11, 2009

tommykins · May 11, 2009

trig (1 Viewer)

ozidolroks

Member

rheyn

Member

Drongoski

Well-Known Member

tommykins

i am number -e^i*pi

tommykins

i am number -e^i*pi

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