trigonometry question 2

**math123456** · December 30th 2008, 06:19 PM

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

**nzmathman** · December 30th 2008, 06:23 PM

If $\cos{x} = \frac{4}{5}$ , then $x = cos^{-1}\left( \frac{4}{5}\right) = 36.9^o$ (acute angle).

Simply substitute this x value into the other equations:

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

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

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

**badgerigar** · December 30th 2008, 07:45 PM

If you need an exact answer, you can use the definition of cos using a triangle.

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)

**Mush** · January 1st 2009, 08:03 PM

Originally Posted by math123456

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

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.

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

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

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

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

$=\frac{2sin(x)cos(x)cos(x)+sin(x)(cos^2(x)-sin^2(x))}{(cos^2(x)-sin^2(x))cos(x)-2sin(x)cos(x)sin(x))}$

**11rdc11** · January 1st 2009, 08:05 PM

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

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

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

Now just plug in and solve

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