prove that sin x/10 + sin 13x/10 =-1/2
huh?
To prove the equation sin(x/10) + sin(13x/10) = -1/2, we will use the identity for the sum of two sines:
sin(A) + sin(B) = 2 * sin((A + B) / 2) * cos((A - B) / 2)
Let's express the equation sin(x/10) + sin(13x/10) as the sum according to the identity:
2 * sin((x/10 + 13x/10) / 2) * cos((x/10 - 13x/10) / 2) = -1/2
Let's simplify the expression inside the sine and cosine functions:
2 * sin(7x/10) * cos(-6x/10) = -1/2
By using the identity sin(-θ) = -sin(θ) and cos(-θ) = cos(θ), we can rewrite the equation as:
2 * sin(7x/10) * cos(6x/10) = -1/2
Now, let's use the double angle identity for sine:
sin(2θ) = 2 * sin(θ) * cos(θ)
By comparing it with our equation, we can rewrite it as:
sin(7x/10) * cos(6x/10) = -1/4
Next, we will use the identity sin(θ) * cos(φ) = (sin(θ + φ) + sin(θ - φ)) / 2:
[(sin(7x/10 + 6x/10) + sin(7x/10 - 6x/10)) / 2] = -1/4
simplifying inside the bracket:
[(sin(13x/10) + sin(x/10)) / 2] = -1/4
Now, we can rewrite this equation as:
(sin(13x/10) + sin(x/10)) / 2 = -1/4
Multiplying both sides of the equation by 2:
sin(13x/10) + sin(x/10) = -1/2
Thus, we have proved that sin(x/10) + sin(13x/10) = -1/2.
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