+0  
 
0
74
3
avatar

i need to find (cosθ)\({^2}\) = 1 in the domain \({-360°≤θ <360°}\)

 

i know i can use the unit circle and quadrantal angles. however, i'm thrown off by the square because i've never encountered a problem like this before? how should i solve it?

 

for instance, (cos 180)​\({^2}\) = -1 isn't possible. however, 180° is a possible answer?

 

the answers in the textbook are -360°, -180°, 0°, and 180°.

 Jan 10, 2019
edited by Guest  Jan 10, 2019

Best Answer 

 #1
avatar+4457 
+2

\(\cos^2(\theta) = 1\\ \text{there are only two values }x \text{ such that }x^2=1\text{, these are}\\ x=1,~x=-1\\ \cos(\theta)=1 \Rightarrow \theta = -360^\circ,~\theta=0^\circ\\ \cos(\theta)=-1 \Rightarrow \theta = -180^\circ,~\theta = 180^\circ\)

.
 Jan 10, 2019
 #1
avatar+4457 
+2
Best Answer

\(\cos^2(\theta) = 1\\ \text{there are only two values }x \text{ such that }x^2=1\text{, these are}\\ x=1,~x=-1\\ \cos(\theta)=1 \Rightarrow \theta = -360^\circ,~\theta=0^\circ\\ \cos(\theta)=-1 \Rightarrow \theta = -180^\circ,~\theta = 180^\circ\)

Rom Jan 10, 2019
 #2
avatar
0

so it's best if i think in terms of \(x\) when facing a quadratic trig/similar trig equations?

Guest Jan 10, 2019

23 Online Users

avatar
avatar