# Thread: tan trig angle formula

1. ## tan trig angle formula

Determine an exact value for the expression:

[sin(pi/3)-cos(5pi/6)]
----------------------
[1-tan(3pi/4)cot(pi/4)]

ok, here are my steps: can someone tell me where my errors are

the answer is supposed to be sqrt3/2

[sin(pi/3)-cos(5pi/6)]
----------------------
[1-tan(3pi/4)cot(pi/4)]
[(sqrt3/2)-(-sqrt3/2)]/[1-(-1)(1)
[sqrt3/2+sqrt3/2]/2
[sqrt9/2]/2
sqrt3

I seem to have lost the 2 in the denominator? Where is my error? thanks

2. Could you elaborate slightly on what your problem is? It looks rather straightforward. Do you have problems calculating which each of the trig functions are? If so, try drawing the unit circle.

3. Originally Posted by skeske1234
I don't understand the tan trig angle formula:

So, how would I do this question?

Determine an exact value for the expression:

[sin(pi/3)-cos(5pi/6)]/[1-tan(3pi/4)cot(pi/4)]

Can you give me all the details and steps please
Just go ahead and calculate the individual trig values! For example, " $\pi/3$" is the same as 60 degrees so you can think of a 60-30 right triangle as half an equilateral triangle. If you take the hypotenuse to be "h" then the leg adjacent to the 60 degree angle has length h/2 and you can find the length of the other leg by using the Pythagorean theorem. Once you have the three lengths, you can get $cos(\pi/3)$ and $sin(\pi/3)$. For $tan(4\pi/4)$ and $cot(\pi/4)$, use the fact that a " $\pi/4-\pi/4$" or "45-45" triangle is isosceles. The "near side" and "opposite" side have the same length.

4. ok, here are my steps: can someone tell me where my errors are

the answer is supposed to be sqrt3/2

[sin(pi/3)-cos(5pi/6)]/[1-tan(3pi/4)cot(pi/4)]
[(sqrt3/2)-(-sqrt3/2)]/[1-(-1)(1)
[sqrt3/2+sqrt3/2]/2
[sqrt9/2]/2
sqrt3

I seem to have lost the 2 in the denominator? Where is my error? thanks

5. Hello, skeske1234!

Here are my steps:

$\frac{\sin\frac{\pi}{3} - \cos\frac{5\pi}{6}} {1-\tan\frac{3\pi}{4}\cot\frac{\pi}{4}}$

. . $=\; \frac{\dfrac{\sqrt{3}}{2}-\left(-\dfrac{\sqrt{3}}{2}\right)} {1-(-1)(1)}$

. . $= \;\frac{\dfrac{\sqrt{3}}{2} +\dfrac{\sqrt{3}}{2}}{2}$

. . $= \;\frac{\left(\dfrac{\sqrt{{\color{red}9}}}{2}\rig ht)}{2}$ . . . nine?

You know this, right? . $\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \;=\;\sqrt{3}$