2 trig questions on ?

[igwun]

1. If sin?=x/2, find cos? in terms of x.

2. Find all ? such that cos?=-1/2

Thanks!

 

[Aladdin]

1. If sin?=x/2, find cos? in terms of x.

2. Find all ? such that cos?=-1/2

Thanks!

sin^2+cos^2=1, Use it

You have to find theta (?) , use the calculator, Actually cos(-1/2)=cos(pi - pi/3)

 

[igwun]

I don't understand how I use sin^2 + cos^2 =1 in the first question.

 

[galactus]

If \(\displaystyle sin({\theta})=\frac{x}{2}\), then \(\displaystyle cos({\theta})=\frac{\sqrt{4-x^{2}}}{2}\)

See from the diagram?. We can use Pythagoras to find the adjacent side. \(\displaystyle \sqrt{2^{2}-x^{2}}=\sqrt{4-x^{2}}\)

Then, \(\displaystyle cos({\theta})\) is adjacent over hypoteneuse. So, we have \(\displaystyle cos({\theta})=\frac{\sqrt{4-x^{2}}}{2}\)


\(\displaystyle cos({\theta})=\frac{-1}{2}\)

\(\displaystyle {\theta}=2C{\pi}+\frac{2\pi}{3}, \;\ {\theta}=2C{\pi}-\frac{2\pi}{3}\)

Let C=0, 1, 2, 3, .......

 

[Deleted member 4993]

I don't understand how I use sin^2 + cos^2 =1 in the first question.

sin(?) = x/2

sin[sup:3dbzv3je]2[/sup:3dbzv3je](?) = x[sup:3dbzv3je]2[/sup:3dbzv3je]/4 .........................(1)

We know

sin[sup:3dbzv3je]2[/sup:3dbzv3je](?) + cos[sup:3dbzv3je]2[/sup:3dbzv3je](?) = 1

cos[sup:3dbzv3je]2[/sup:3dbzv3je](?) = 1 - sin[sup:3dbzv3je]2[/sup:3dbzv3je](?) ..... Using (1) we get

cos[sup:3dbzv3je]2[/sup:3dbzv3je](?) = 1 - x[sup:3dbzv3je]2[/sup:3dbzv3je]/4 = ( 4 - x[sup:3dbzv3je]2[/sup:3dbzv3je])/4

cos(?) = ±?( 4 - x[sup:3dbzv3je]2[/sup:3dbzv3je])/2

What was so hard about that ??!! - did you even try with paper/pencil??