Answers>Maths>IB>Article

Given that sin(x) + cos(x) = 2/3, find cos(4x)

It is not clear what to do when starting, but we realise that by making both sides to the power of two, will lead to have an expression containing sin^2(x) + cos^2(x), which is equal to one (which will probably make things easier):
sin^2(x) + cos^2(x) + 2sin(x)cos(x) = 4/91 + 2sin(x)cos(x) = 4/9 2sin(x)cos(x) = -5/9
We also know thanks to double angle identities that 2sin(x)cos(x) is just sin(2x) so we just substitute:
sin(2x) = -5/9
Now we have an expression in terms of sine, but we want it in terms of cos. We take a look at the formulas for double angles (double angle identities) and find out that cos(2k)= 1-2sin^2(k). It is fairly straight forward to solve from this point but to make it simpler, it is useful to make k=2x. So thatcos(4x) = 1-2sin^2(2x)
We know what sin(2x) is socos(4x) = 1- 2* (25/81)cos(4x) = 31/81

Answered by Felipe R. Maths tutor

3604 Views

See similar Maths IB tutors

Related Maths IB answers

All answers ▸

Consider f (x) = logk (6x - 3x 2 ), for 0 < x < 2, where k > 0. The equation f (x) = 2 has exactly one solution. What is the value of k?


Given that y = -16x2​​​​​​​ + 160x - 256, find the value of x giving the maximum value of y, and hence give this maximum value of y.


Which are the difference between polar and coordinate complex numbers?


What is the limit for this function as x approaches 0? y(x)=(cos x)^(1/sin x)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences