Question

Show how

[sin((x+0.5)theta) - sin((x-0.5)theta)] / [2sin(theta/2)] = cos(x theta)

Please show all steps clearly.

Please do not overlook the theta in two places in the numerator.

Answer 1

Solution

sin((x+0.5)theta)-sin((x-0.5)theta)/[2sin(theta/2)]=cos(xtheta)

Solve left hand side

We know a formula

sinA-sinB=2cos((A+B)/2)×sin((A-B)/2)

so, we need to apply this formula but first identity A and B

in this case, A=(x+0.5)theta and B=(x-0.5)theta

So, we solve (A+B)/2 and (A-B)/2 and then put in the formula

So, (A+B)/2=[(x+0.5) theta+(x-0.5)theta]/2

take theta common above then simplify

=theta (x+0.5+x-0.5)/2=theta(2x)/2

Now 2 will be divided by 2 because 2 is in numerator and as well denominator

So, we get

(A+B)/2=xtheta

Now, (A-B)/2=[(x+0.5)theta-(x-0.5)theta]/2

=theta(x+0.5-x+0.5)/2=theta(x-x+0.5+0.5)/2

=theta(1)/2=theta/2

So, (A-B)/2=theta/2

Now, put these in formula we get

Sin((x+0.5)theta)-sin((x-0.5)theta)=2cos(xtheta)×sin(theta/2)

now, put this value of sin((x+0.5)theta)-sin((x-0.5)theta) in the left hand side we get

2cos(xtheta)×sin(theta/2)/2sin(theta/2)

here, we can see numerator and denominator has 2sin(theta/2) so, they will cancel out and gives 1

we get

cos(xtheta)×1

and cos(xtheta)

Therefore, left hand side=right hand side

Hence, proved