+0

# Finding sin and tan given cos?

0
1160
4

if cos(x)=1/2, how do i find sin and tan?

Guest Apr 7, 2017

#1
+7340
+3

Here is a drawing of what cos(x)= 1/2 really means:

We can find sin(x) just using the Pythagorean Theorem.

$$(\frac12)^2+(\sin (x))^2=1^2 \\~\\ \mathbf{sin(x)}=\sqrt{1-\frac14}=\sqrt{\frac34}\mathbf{=\frac{\sqrt3}{2}}$$

tan = sin / cos, so...

tan(x) = sin(x) / cos(x) = $$\frac{\sqrt3}{2}\div\frac{1}{2}=\frac{\sqrt3}{2}\cdot\frac{2}{1}\mathbf{=\sqrt3}$$

hectictar  Apr 8, 2017
edited by hectictar  Apr 8, 2017
#1
+7340
+3

Here is a drawing of what cos(x)= 1/2 really means:

We can find sin(x) just using the Pythagorean Theorem.

$$(\frac12)^2+(\sin (x))^2=1^2 \\~\\ \mathbf{sin(x)}=\sqrt{1-\frac14}=\sqrt{\frac34}\mathbf{=\frac{\sqrt3}{2}}$$

tan = sin / cos, so...

tan(x) = sin(x) / cos(x) = $$\frac{\sqrt3}{2}\div\frac{1}{2}=\frac{\sqrt3}{2}\cdot\frac{2}{1}\mathbf{=\sqrt3}$$

hectictar  Apr 8, 2017
edited by hectictar  Apr 8, 2017
#2
+92856
+2

cos    = x / r      sin  = y / r   and tan  = y / x

We know x and r and we need to find y =  sqrt (r^2 - x^2)  = sqrt (2^2 - 1^2) =

sqrt (4 - 1)   =  sqrt (3)

So

sin (x)  = y/r =  sqrt (3) / 2       and tan (x) = y/x   =  sqrt (3)  / 1  = sqrt (3)

CPhill  Apr 8, 2017
#3
+3478
0

sqrt3

tertre  Apr 8, 2017
#4
+310
+2

You forgot a solution. If cos(x)=1/2 then sin(x)=(3/4)1/2 OR sin(x)=-(3/4)1/2

That means tan(x)=-(31/2) OR tan(x)=31/2

Ehrlich  Apr 8, 2017