Question

Prove that sin(270+A) +cosA + cos(180+A)- sin(270-A)=0

Answer 1

cos(A)+sin(270°+A)-sin(270°-A)+cos(180°... 0?  

Note that sin (90° + A) = sin (90° - A) = cos A ;  

and sin (180° ± A) = - sin A -- similarly for cos.  

So:  

sin (270° + A) = - sin (90° + A) = cos A  

sin (270° - A) = - sin (90° - A) = cos A  

Their difference is zero.  

We are left with cos A + cos (180° + A) , = cos A - cos A  

The result is 0, as expected.

Answer 2

sin (270+A) =-cos A cos (180+A)= -cosA
sin (270-A) =-cos A
therefore left hand side,

sin (270+A) +cosA + cos (180+A) - sin(270-A)
= -cosA + cosA - cosA + cosA
= 0

hence proved
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