Question

How do you use the sum and difference identity to evaluate `sin((7pi)/12)`?

Answer 1

`sqrt(2 + sqrt3)/2`

Explanation:

Trig unit circle -->
`sin ((7pi)/12) = sin (pi/12 + pi/2) = cos (pi/12)`
Find `cos (pi/12)` by using the trig identity:
`cos 2a = 2cos^2 a - 1`
`cos ((2pi)/12) = cos (pi/6) = sqrt3/2 = 2cos^2 (pi/12) - 1`
`2cos^2 (pi/12) = 1 + sqrt3/2 = (2 + sqrt3)/2`
`cos^2 (pi/12) = (2 + sqrt3)/4`
`cos (pi/12) = +- sqrt(2 + sqrt3)/2.`
Since cos (pi/12) is positive, take the positive answer.
Finally,
`sin ((7pi)/12) = cos (pi/12) = sqrt(2 + sqrt3)/2`

Answer 2

`sin(7pi/12)=(sqrt6+sqrt2)/4`.

Explanation:

We have, `sin(pi-theta)=sintheta`.

`rArr sin(7pi/12)=sin (pi-5pi/12)=sin5pi/12`.

Now, `sin(A+B)=sinAcosB+cosAsinB`.

Taking, `A=pi/4, and, B=pi/6`,

we have, `A+B=pi/4+pi/6=3pi/12+2pi/12=5pi/12`.

`:. sin5pi/12=sin(pi/4+pi/6)`.

`=sin(pi/4)cos(pi/6)+cos(pi/4)sin(pi/6)`.

`=1/sqrt2*sqrt3/2+1/sqrt2*1/2=(sqrt3+1)/(2sqrt2)=(sqrt6+sqrt2)/4`.

Finally, `sin(7pi/12)=(sqrt6+sqrt2)/4`.

Just to match this Answer with that furnished by Respected Nghi N. , we see that,

`(sqrt3+1)/(2sqrt2)=sqrt{((sqrt3+1)/(2sqrt2))^2}=sqrt{((3+1+2sqrt3)/8)`

`=sqrt((4+2sqrt3)/8)=sqrt{(2(2+sqrt3))/8}=sqrt{(2+sqrt3)/4}=sqrt(2+sqrt3)/2`

Enjoy Maths.!