1. ## [SOLVED]Finding Trig Values

Question is:

If A is an acute angle and CosA=4/5, find the values of,

a) sin2A

b) Sin3A

c) Tan3A
I sort of have an understanding with this question, but i dont really get it, especially part b) and part c). With a solution, can someone also leave an explanation to what they did, so i can try to understand what you did? Thanks.

2. Originally Posted by iMan_07
Question is:

I sort of have an understanding with this question, but i dont really get it, especially part b) and part c). With a solution, can someone also leave an explanation to what they did, so i can try to understand what you did? Thanks.
You know from basic trigonometry, that if you consider a right angled triangle, then $\displaystyle cos(A) = \frac{a}{h}$, where A is the angle you are considering, a is the length of the adjecent side, and h is the length of the hypotenuse.

In this case, our triangle has adjacent side length 4, and hypotenuse of 5. Perhaps it would help if you sketched it!

Using pythagorus, you can work out the length of the side opposite the triangle.

$\displaystyle o^2 = h^2 - a^2$

$\displaystyle = 5^2-4^2$

$\displaystyle = 25-16 = 9$

$\displaystyle \therefore o = \sqrt{9} = 3$

$\displaystyle o = 3$

So it's a 3,4,5 triangle! Again, from basic trigonometry, you can calculate $\displaystyle sinA$:

$\displaystyle sin(A) = \frac{o}{h} = \frac{3}{5}$.

So now you know sinA and cosA.

a) You need $\displaystyle sin2A$. Remember $\displaystyle sin2A = 2sinAcosA$. You know both!

b) You need $\displaystyle sin3A$. Remeber that $\displaystyle sin(3A) = sin(2A+A) = sin2AcosA+cos2AsinA$. And remember that $\displaystyle cos(2A) = (cosA)^2 - (sinA)^2$.

c) You need $\displaystyle tan(3A)$. Remember $\displaystyle tan(3A) = \frac{sin(3A)}{cos(3A)}$. You worked out $\displaystyle sin(3A)$ before, you just need $\displaystyle cos(3A)$. Use the same concept of:

$\displaystyle cos(3A) = cos(2A+A) = cos2AcosA-sin2AsinA$ and $\displaystyle cos(2A) = (cosA)^2-(sinA)^2$

Good luck.

3. Thank You soo much, wonderful explanation.