If x is an acute angle and cosx=4/5, find the values of sin2x, sin3x, tan3x.

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- Dec 30th 2008, 06:19 PMmath123456trigonometry question 2
If x is an acute angle and cosx=4/5, find the values of sin2x, sin3x, tan3x.

- Dec 30th 2008, 06:23 PMnzmathman
If $\displaystyle \cos{x} = \frac{4}{5}$, then $\displaystyle x = cos^{-1}\left( \frac{4}{5}\right) = 36.9^o$ (acute angle).

Simply substitute this x value into the other equations:

$\displaystyle \sin{(2 \times 36.9^o)} = $?

$\displaystyle \sin{(3 \times 36.9^o)} = $?

$\displaystyle \tan{(3 \times 36.9^o)} = $? - Dec 30th 2008, 07:45 PMbadgerigar
If you need an exact answer, you can use the definition of cos using a triangle.

Attachment 9418

Pythagoras' theorem can help you find sin(x). Then you can find sin(2x) and cos(2x) using the double angle formulas and then use sin(3x) = sin(x+2x) and cos(3x) = cos(2x+x) - Jan 1st 2009, 08:03 PMMush
cos(x) = adjacent/hypotenuse.

So adjacent = 4

Hypotenuse = 5.

You can use pythagorus to work out the opposite side, and the use that to find sin(x) = opposite/hypotenuse.

$\displaystyle sin2x = 2sin(x)cos(x)$

$\displaystyle sin(3x) = sin(2x+x) = sin(2x)cos(x) + sin(x)cos(2x) $

$\displaystyle = 2sin(x)cos(x)cos(x)+sin(x)(cos^2(x)-sin^2(x)) $

$\displaystyle tan(3x) = \frac{sin(3x)}{cos(3x)} = \frac{sin(2x+x)}{cos(2x+x)}=\frac{sin(2x)cos(x) + sin(x)cos(2x)}{cos(2x)cos(x)-sin(2x)sin(x)} $

$\displaystyle =\frac{2sin(x)cos(x)cos(x)+sin(x)(cos^2(x)-sin^2(x))}{(cos^2(x)-sin^2(x))cos(x)-2sin(x)cos(x)sin(x))}$ - Jan 1st 2009, 08:05 PM11rdc11