# Thread:

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 $cos(A) = \frac{a}{h}$, 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.

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

$= 5^2-4^2$

$= 25-16 = 9$

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

$o = 3$

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

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

So now you know sinA and cosA.

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

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

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

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

Good luck.

3. Thank You soo much, wonderful explanation.
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