Expert: Alon Mandes - 9/5/2009

Question
sin(2A)=2sinA cosA
 =(2sinA cos^2A)/cosA
 =2tanA/sec^2A
 =2tanA/1+tan^2A

please explain it

Answer
Let's decompose the answer into steps :
Step 1 : According to the law of trigonometry : sin(2A)=2sin(A)cos(A) . Let's prove it :
        We know that : sin(a+b)=sin(a)cos(b)+sin(b)cos(a). Let's substitute b=a , we get :
        sin(a+a)=sin(a)cos(a)+sin(a)cos(a)  -> sin(2a)=2sin(a)cos(a) .
Step 2 : sin(2A)=[2sin(A)(cos(A))²]/cos(A) . Let's prove it :
        sin(2a)=2sin(a)cos(a) --> sin(2a)=[2sin(a)cos(a)]*1 -->
        sin(2a)=[2sin(a)cos(a)]*[cos(a)/cos(a)] --> sin(2a)=[2sin(a)cos(a)²]/cos(a) .
Step 3 : sin(2A)=2tan(A)/sec(A)² . This is true because as we know tang(a)=sin(a)/cos(a) &
        sec(a)=1/cos(a) . Therefore :
        sin(2a)=[2sin(a)cos(a)²]/cos(a)=[2sin(a)/cos(a)][cos(a)²].
        It's clear that cos(a)²=1/sec(a)² . Hence :
        sin(2A)=2tan(A)/sec(A)² .
Step 4 : sin(2A)=2tanA/1+tan^2A . All we need to prove here is that sec(a)²=1+tan(a)² .
        Let's do it :
        1+tan(a)²=1 + sin(a)²/cos(a)² = [cos(a)²+sin(a)²]/cos(a)² =
        [cos(a)²+sin(a)²]*[1/cos(a)²]=[cos(a)²+sin(a)²]*[sec(a)²] . we know from trigonometry
        that cos(a)²+sin(a)²=1 , this gives us : 1+tan(a)²=1*[sec(a)²]= sec(a)² . Q.E.D

Alon.