Using the identity cos(A+B)= cosAcosB-sinAsinB, prove that cos2A=1-2sin^2A.

Use cos(A+B)=cosAcosB-sinAsinB and let A=B so cos(A+A)=cosAcosA-sinAsinA this means cos(2A)=cos2A-sin2A and since cos2A+sin2A=1, cos2A=1-sin2A. Therefore, by subbing cos2A=1-sin2A into cos(2A)=cos2A-sin2A, we get cos(2A)=1-sin2A-sin2A=1-2sin2A.

RF
Answered by Rebecca F. Maths tutor

21918 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

A curve has equation y = x^3 - 48x. The point A on the curve has x coordinate -4. The point B on the curve has x coordinate - 4 + h. Show that that the gradient of the line AB is h^2 - 12h.


How would I find the approximate area enclosed by the expression e^x*sin(x)*x^3 on an infinite scale?


Chris claims that, “for any given value of x , the gradient of the curve y=2x^3 +6x^2 - 12x +3 is always greater than the gradient of the curve y=1+60x−6x^2” . Show that Chris is wrong by finding all the values of x for which his claim is not true.


Differentiate x^x


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning