How do you use the sum and difference identity to evaluate sin((7pi)/12)sin(7π12)?

2 Answers
Aug 20, 2016

sqrt(2 + sqrt3)/22+32

Explanation:

Trig unit circle -->
sin ((7pi)/12) = sin (pi/12 + pi/2) = cos (pi/12)sin(7π12)=sin(π12+π2)=cos(π12)
Find cos (pi/12)cos(π12) by using the trig identity:
cos 2a = 2cos^2 a - 1cos2a=2cos2a1
cos ((2pi)/12) = cos (pi/6) = sqrt3/2 = 2cos^2 (pi/12) - 1cos(2π12)=cos(π6)=32=2cos2(π12)1
2cos^2 (pi/12) = 1 + sqrt3/2 = (2 + sqrt3)/22cos2(π12)=1+32=2+32
cos^2 (pi/12) = (2 + sqrt3)/4cos2(π12)=2+34
cos (pi/12) = +- sqrt(2 + sqrt3)/2.cos(π12)=±2+32.
Since cos (pi/12) is positive, take the positive answer.
Finally,
sin ((7pi)/12) = cos (pi/12) = sqrt(2 + sqrt3)/2sin(7π12)=cos(π12)=2+32

Aug 20, 2016

sin(7pi/12)=(sqrt6+sqrt2)/4sin(7π12)=6+24.

Explanation:

We have, sin(pi-theta)=sinthetasin(πθ)=sinθ.

rArr sin(7pi/12)=sin (pi-5pi/12)=sin5pi/12sin(7π12)=sin(π5π12)=sin5π12.

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

Taking, A=pi/4, and, B=pi/6A=π4,and,B=π6,

we have, A+B=pi/4+pi/6=3pi/12+2pi/12=5pi/12A+B=π4+π6=3π12+2π12=5π12.

:. 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.!