A community for students.
Here's the question you clicked on:
 0 viewing
amydh
 10 months ago
sin^2 theta  cos^2 theta = 1
amydh
 10 months ago
sin^2 theta  cos^2 theta = 1

This Question is Closed

ParthKohli
 10 months ago
Best ResponseYou've already chosen the best response.2\[\left(\cos^2 \theta  \sin^2 \theta\right) = 1\]\[\implies \cos(2\theta) = 1\]\[\implies \cos(2\theta) = 1\]Or use either of the two identities:\[\cos^2 \theta = 1  \sin^2 \theta\]\[\sin^2 \theta = 1 \cos^2 \theta\]

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.2or ... ANOTHER WAY! \[\sin^2(\theta)\cos^2(\theta)=1 \] Recall \[\sin^2(\theta)+\cos^2(\theta)=1\] So add \[\cos^2(\theta)\] on both sides to use the identity that is being recalled \[\sin^2(\theta)+\cos^2(\theta)\cos^2(\theta)=1+\cos^2(\theta) \] Now use the identity \[1\cos^2(\theta)=1+\cos^2(\theta)\] \[2=2\cos^2(\theta)\] \[1=\cos^2(\theta) \] which results into two equations

ParthKohli
 10 months ago
Best ResponseYou've already chosen the best response.2Just add those two equations to get \(2\sin^2 \theta = 0\).

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.2oh yeah that is pretty too

mathslover
 10 months ago
Best ResponseYou've already chosen the best response.0This problem can be done in several ways. I was going to write some of them here, but I guess, they will all be actually similar to each other. But @myininaya and @ParthKohli have done well to find 2 of them.

myininaya
 10 months ago
Best ResponseYou've already chosen the best response.2parth did one way at first then he found another way off what i was doing

ParthKohli
 10 months ago
Best ResponseYou've already chosen the best response.2why are you typing like satellite does
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.