# exact value question

• December 18th 2007, 11:34 AM
Hydralore
exact value question
If cos (-) = 4/5 , 0 < (-) < pie/2

find the exact value of

(a) sin 2(-)
(b) sin 4(-)

given that tan x = square root of 11 all over 3, o < x < pie/2 , find the exact value of sin x

If anyone could help me with these questions id be gratefull :)

edit : damn.. can anyone help me show that sin (2x + x) = 3sin x - 4sin(to the power of 3)x
• December 18th 2007, 11:41 AM
bobak
$\cos \theta = \frac{4}{5}$

using $\sin^2 \theta + \cos^2 \theta = 1$ or and appropriate triangle calculate the exact value of $\sin \theta$.

you should now that $\sin 2 \theta = 2\sin \theta \cos \theta$
this should be an easy calculation.

similarly $\sin 4 \theta = 2\sin 2 \theta \cos 2 \theta$
should be able to take it form there
• December 18th 2007, 11:44 AM
Hydralore
Thanks,

i also thought i could do another question but i need a little help.

show that sin (2x + x) = 3 sin x - 4 sin^3x
• December 18th 2007, 11:54 AM
bobak
Quote:

show that $\sin (2\theta + \theta) = 3 \sin \theta - 4 \sin^3 \theta$
starting on the LHS using the compound angle formula
$\sin (2\theta + \theta) = \sin 2 \theta \cos \theta + \cos 2 \theta \sin \theta$

you know that $\sin 2\theta = 2\sin \theta \cos \theta$
and you should know $\cos 2\theta = cos^2 \theta - \sin^2 \theta$

you may need to use $\cos^2 \theta = 1 - \sin^2 \theta$
• December 18th 2007, 03:00 PM
Krizalid
Quote:

Originally Posted by Hydralore
can anyone help me show that sin (2x + x) = 3sin x - 4sin(to the power of 3)x

$\sin 3x = \text{Im} \left( {e^{ix} } \right)^3 = \text{Im} \left( {\cos x + i\sin x} \right)^3 .$

Now

$\left( {\cos x + i\sin x} \right)^3 = \cos ^3 x + 3i\sin x\cos ^2 x - 3\sin ^2 x\cos x - i\sin ^3 x.$

And finally

$\sin 3x = 3\sin x(1 - \sin ^2 x) - \sin ^3 x = 3\sin x - 4\sin ^3 x\,\blacksquare$
• December 21st 2007, 04:39 PM
JaneBennet
Quote:

Originally Posted by Hydralore
given that tan x = square root of 11 all over 3, o < x < pie/2 , find the exact value of sin x

Calculate the hypotenuse using Pythagoras’ theorem and use $\mbox{sine} = \frac{\mbox{opposite}}{\mbox{hypotenuse}}$.
• December 24th 2007, 10:58 PM
quq
Quote:

Originally Posted by JaneBennet
Calculate the hypotenuse using Pythagoras’ theorem and use $\mbox{sine} = \frac{\mbox{opposite}}{\mbox{hypotenuse}}$.

hypotenuse = $\sqrt{20}$

sin(x) = $\sqrt{ \frac{11} {20} }$

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