prove that sin x/10 + sin 13x/10 =-1/2
2 answers
huh?
sin x/10 + sin 13x/10 =-1/2
one of our trig identities is
sinA + sinB = 2sin ((A+B)/2) cos((A-B)/2)
so yours is
2sin((x/10+13x/10)/2)cos((x/10-13x/10)/2) = -1/2
sin(7x/10)cos(-3x/5) = -1/4
sin(7x/10)cos(3x/5) = -1/4 , cos(-x) = cosx
sin(7x/10)cos(6x/10) = -1/4
Amazingly, this is actually true for x = π , but the graph shows it is not true for "many" other values of x
Try any value of x, make sure your calculator is set to RAD
All you need is ONE exception and your identity is not true.
www.wolframalpha.com/input/?i=sin(7x%2F10)cos(6x%2F10)+%3D+-1%2F4
one of our trig identities is
sinA + sinB = 2sin ((A+B)/2) cos((A-B)/2)
so yours is
2sin((x/10+13x/10)/2)cos((x/10-13x/10)/2) = -1/2
sin(7x/10)cos(-3x/5) = -1/4
sin(7x/10)cos(3x/5) = -1/4 , cos(-x) = cosx
sin(7x/10)cos(6x/10) = -1/4
Amazingly, this is actually true for x = π , but the graph shows it is not true for "many" other values of x
Try any value of x, make sure your calculator is set to RAD
All you need is ONE exception and your identity is not true.
www.wolframalpha.com/input/?i=sin(7x%2F10)cos(6x%2F10)+%3D+-1%2F4