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

Question

Brooklyn Anderson · Accepted Answer

#sqrt(2 + sqrt3)/2# 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#

James Adkins · Answer

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

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#. In order to compare this response with the one provided by Respected Nghi N., we observe 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# Have fun with math!

Violet Anderson · Answer

undefined To evaluate sin((7π)/12) using the sum and difference identity, we first express (7π)/12 as the sum or difference of angles whose sine values we know. In this case, we can rewrite (7π)/12 as (π/4) + (π/3). Then, we apply the sum identity for sine, which states that sin(A + B) = sinA * cosB + cosA * sinB. By substituting π/4 for A and π/3 for B, we can calculate sin((7π)/12) as sin(π/4) * cos(π/3) + cos(π/4) * sin(π/3). Simplifying these trigonometric functions gives us (√2/2) * (1/2) + (√2/2) * (√3/2). Further simplifying, we get (√2/4) + (√6/4), which equals (√2 + √6)/4. Therefore, sin((7π)/12) = (√2 + √6)/4.