# trigonometry question 2

• Dec 30th 2008, 06:19 PM
math123456
trigonometry question 2
If x is an acute angle and cosx=4/5, find the values of sin2x, sin3x, tan3x.
• Dec 30th 2008, 06:23 PM
nzmathman
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 PM
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 PM
Mush
Quote:

Originally Posted by math123456
if x is an acute angle and cos x=4/5, find the exact values of sin2a, sin3a and tan3a.

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 PM
11rdc11
Quote:

Originally Posted by math123456
if x is an acute angle and cos x=4/5, find the exact values of sin2a, sin3a and tan3a.

Ok since $\displaystyle \cos{x} = \frac{4}{5}$ is an acute ww can concur that sin is positive as well

Using pythagorean theorem we see that $\displaystyle \sin{x} = \frac{3}{5}$

Now $\displaystyle \sin{2x} = 2\sin{x}\cos{x}$

Now just plug in and solve