Question

Using the identity Cos(A+B)=CosACosB+SinASinB
Prove that Cos2A=1-2Sin^2A

Answer 1

put B=A
cos2A=cosA.cosA+sinA.sinA

Answer 2

Sorry top bit should be Cos(A+B)=CosACosB-SinASinB

Answer 3

sry then also do the same

Answer 4

(cosA)^2+sinA)^2=1-(sina)^2-(sina)^2

Answer 5

\(\large {\cos 2A = \cos (A+A) = \cos A \cos A - \sin A \sin A \\
\implies \cos 2A = \cos^2 A - \sin^2 A \\
\implies \cos 2A = (1-\sin^2 A ) - \sin^2 A ... \quad \textbf{as : } \space \cos^2 A = 1 - \sin^2 A \\
\implies \cos 2A = 1 - 2\sin^2 A ... \boxed{\textbf{Proved}}} \)

Answer 6

Isnt it\[\cos ^{2}A-1=1-\sin ^{2}A\] ?

Answer 7

No, see : \(\large \cos^2 A + \sin^2 A= 1\)

So, \(\large \cos^2 A = 1 - \sin^2 A\)

\(\implies \large{\cos^2 A - 1 = -\sin^2 A}\)

Answer 8

Ok I understand, my maths formula sheet was a bit confusing.
Thanks for the help!

Answer 9

You're welcome. Best of Luck and let me know if you have any more questions any time.