Trigonometric Double Angle Formulas

[stevo_evo_22]

Hi everyone!

Here is the question I am stuck on:

Find the value of theta and state any restrictions on the value of theta: {theta:cos2theta=cos^2theta}, 0_<_theta_<_2pi.

(Sorry if you can't understand the question, i'm not used to using math tags)

I have solved for costheta and worked it out to be 1 (or at least I think) but when it come to giving the value of theta, i only get 0 and 2pi. The book's answer also includes 3pi/2. Where have I gone wrong?

Thanks in advance,
Steven

 

[royhaas]

You ignored $\displaystyle \cos(\theta) = - 1$.

 

[soroban]

Hello, Steven!

Roy is absolutely correct . . .

Find the value of $\displaystyle \theta$ and state any restrictions on $\displaystyle \theta$

. . \(\displaystyle \cos2\theta \:=\:\cos^2\!\theta},\qquad 0\,\leq \,\theta \,<\,2\pi\)

$\displaystyle \text{Use the identity: }\:\cos2\theta \:=\:2\cos^2\!\theta - 1$

$\displaystyle \text{The equation becomes: }\:2\cos^2\!\theta - 1 \:=\:\cos^2\!\theta \quad\Rightarrow\quad \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \cos\theta \:=\:\pm1$


. . $\displaystyle \text{Then: }\:\cos\theta\:=\:1 \quad\Rightarrow\quad \theta \:=\:0$

. . $\displaystyle \text{And: }\:\cos\theta \:=\:\text{-}1 \quad\Rightarrow\quad \theta \:=\: \pi$

Edit: corrected my (really stupid) errors.
Thanks for pointing them out, masters!
.

 

[stevo_evo_22]

Of course! Thankyou to both of you! I will remember next time about square-rooting cosine! Thanks!!!!