Test: Basic Trigonometric Formula

Cos68°cos8°+sin68°sin8°

We know, 

cosAcosB+sinAsinB=cos(A-B)

Cos(68-8)

Cos60°

=1/2

cos(a-b) = Cos a Cos b + Sin a Sin b

Cos(130-40)=cos90=0    

If sin A = 1 / ( 10 ) ^{( 1 / 2 )}and sin B = 1 / ( 5 ) ^{( 1 / 2 )} then cos A = ( 1 - sin ² A ) ^{( 1 / 2 )}

= cos A = 3 / ( 10 ) ^{( 1 / 2 )} ,similarly ,cos B = ( 1 - sin² B )^{(1 / 2 )}

= cos B = 2 / ( 5 ) ^{( 1 / 2 )}

= sin ( A + B ) =  sin ( pi / 4 )

= sin A cos B  +  sin B cos A  = sin ( pi / 4 ) 

Sin A Cos B + Cos A SinB =Sin( A + B) 

 = Sin(120+60)  = Sin (180) = 0

sin (105^º) = sin (90^º + 15^º) = cos 15⁰

SinA+cosA=√2(sinA/√2+cosA/√2)
=√2(sinAcosπ/4+cosAsinπ/4)
=√2sin(A+π/4)
by using sin(A+B) formula

Sin(60+a)cos(30-b)+ cos(60+a)sin(30-b) =sin(60+a+30-b) sin(a)cos(b) + cos(a)sin(b)= sin(a+b) =sin(90+《a-b》) =cos(a-b)

It is equal to SinACosB + CosASinB = cos(A+B)

                  hence, = Sin(45-A+45-B)

                             = Sin(90-(A+B))

                             = Cos(A+B)

Sin(n+1)ASin(n+2)A + Cos(n+1)ACos(n+2)A =Cos (n+1)ACos(n+2)A + Sin(n+1)ASin(n+2)A =Cos{A(n+2-n-1)} =cos (A.1) =cos A

Cos(π/4-x)cos (π/4-y) - sin (π/4-x) sin(π/4-y)

= CosA*Cos B - Sin A*Sin B =

                                      Cos (A+B)

=cos(π/4-x+π/4-y)

=cos(π/2-x-y)

=cos{π/2 - (x+y)}

= sin(x+y)

cos 130cos 40+sin 130sin 40 = cos (130-40) = cos 90 =0