Question

sin 2A = cos (go-2A)
take
A =
18
Please solve this example​

Answer 1

Answer:

Sin2a=cos(a-18)

Sin2a=cos(a-18)[SinA=cos (90-A)]

Sin2a=cos(a-18)[SinA=cos (90-A)]Cos(90-2a)=cos (a-18)

Sin2a=cos(a-18)[SinA=cos (90-A)]Cos(90-2a)=cos (a-18)90-2a=a-18

Sin2a=cos(a-18)[SinA=cos (90-A)]Cos(90-2a)=cos (a-18)90-2a=a-1890+18=a+2a

Sin2a=cos(a-18)[SinA=cos (90-A)]Cos(90-2a)=cos (a-18)90-2a=a-1890+18=a+2a108=3a

Sin2a=cos(a-18)[SinA=cos (90-A)]Cos(90-2a)=cos (a-18)90-2a=a-1890+18=a+2a108=3aa=108/3

Sin2a=cos(a-18)[SinA=cos (90-A)]Cos(90-2a)=cos (a-18)90-2a=a-1890+18=a+2a108=3aa=108/3a=36°

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Sin 2A = cos (90-2A )

sin 2A = sin 2A

hence proved