Tirgonometric Identities and Equations

Algebra
Tutor: None Selected Time limit: 1 Day

Rewrite g(x) = 5 cos (2x - pi/4) - 3 as, k(x), a function of sine. Please show all work. 

Jun 4th, 2015

Thank you for the opportunity to help you with your question!

cosx=sin(x+pi/2)

so 5 cos (2x - pi/4) - 3 =5cos(2(x-pi/8))-3=5sin(2(x-pi/8+pi/2))-3=5sin(2(x+3pi/8))-3

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 4th, 2015

Shouldn't it be: 5 cos (2(x - pi/8)) - 3 = 5 sin (2(x - pi/8) + pi/2) - 3 = 5 sin(2(x + pi/8)) - 3 ?


Jun 4th, 2015

yes you are correct! Can you tell me where I got wrong to enforce the memory?

I know you saw "2(x - pi/8)" as a whole and then added pi/2, that is the correct step!

Jun 4th, 2015

Yes, I treated "2(x - pi/8)" as X, and I think you just missed adding a right parenthesis upon converting from cos to sin. Thank you for your quick response and for solving the 2 problems I posted. One more clarification I'd like to ask for -- how did you get cos x = sin (x+pi/2)? My book does not have it. It has cos(pi/2 - theta) = sin theta

Jun 4th, 2015

Here is the clarification I asked for: how did you get cos x = sin (x+pi/2)? My book does not have it. It has cos(pi/2 - theta) = sin theta

Jun 4th, 2015

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