What are all values for A between 0-360 degrees? sin^2(A/2)=cos^2A

2 Answers
maganbhai P.
Apr 12, 2018

#A=60^circ,180^circ,300^circ.#

Explanation:

Here,

#sin^2(A/2)=cos^2A#

#(1-cosA)/2=cos^2A#

#1-cosA=2cos^2A#

#2cos^2A+cosA-1=0#

#2cos^2A+2cosA-cosA-1=0#

#2cosA(cosA+1)-1(cosA+1)=0#

#(cosA+1)((2cosA-1)=0#

#cosA+1=0 or 2cosA-1=0#

#cosA=-1 or cosA=1/2#,where,# A in [0,360)#

We have two options:

#(1)cosA=-1=>A=180^circ#

#(2)cosA=1/2=>A=60^circ,300^circ#

Hence,

#A=60^circ,180^circ,300^circ.#

Somebody N.
Apr 12, 2018

#color(blue)(A=60^@, 180^@,300^@)#

Explanation:

Identity:

#color(red)bb(sin^2(x/2)=1/2(1-cosx))#

#sin^2(A/2)=cos^2A#

Using identity:

#1/2(1-cosA)=cos^2A#

Simplifying:

#1/2-1/2cosA-cos^2A=0#

#2cos^2A+cosA-1=0#

Let #u=cosA#

#2u^2+u-1=0#

Factor:

#(2u-1)(u+1)=0=>u=-1 and u=1/2#

#:.#

#cosA=-1 and cosA=1/2#

#A=arccos(cosA)=arccos(-1)=>A=180^@#

#A=arccos(cosA)=arccos(1/2)=>A=60^@, 300^@#

Solutions:

#color(blue)(A=60^@, 180^@,300^@)#

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