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Question
Prove that 9π8−94sin−1(13)=94sin−1(2√23)
Prove that 9π8−94sin−1(13)=94sin−1(2√23)
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Solution
Given
9π8−94sin−1(13)=94sin−1(2√23)⇒9π8=94sin−1(13)+94sin−1(2√23)⇒9π8=94[sin−1(13)+sin−1(2√23)]RHS=94[sin−1(13)+sin−1(2√23)]=94[sin−1{13√1−(2√23)2+2√23√1−(13)2}]
[∵ sin−1x+sin−1y=sin−1(x√1−y2+y√1−x2)]=94[sin−1(13×13+2√23×2√23)]=94[sin−1(19+89)]=94sin−1(1)=94×π2=9π8=LHS.
Alternative method
LHS=9π8−94sin−1(13)=94(π2−sin−1(13))=94(cos−1(13))[∵ sin−1x+cos−1x=π2]=94sin−1√1−(12)2 [∵cos−1x=sin−1√1−x2]=94sin−1√1−19=94sin−12√23=RHS
Given
9π8−94sin−1(13)=94sin−1(2√23)⇒9π8=94sin−1(13)+94sin−1(2√23)⇒9π8=94[sin−1(13)+sin−1(2√23)]RHS=94[sin−1(13)+sin−1(2√23)]=94[sin−1{13√1−(2√23)2+2√23√1−(13)2}]
[∵ sin−1x+sin−1y=sin−1(x√1−y2+y√1−x2)]=94[sin−1(13×13+2√23×2√23)]=94[sin−1(19+89)]=94sin−1(1)=94×π2=9π8=LHS.
Alternative method
LHS=9π8−94sin−1(13)=94(π2−sin−1(13))=94(cos−1(13))[∵ sin−1x+cos−1x=π2]=94sin−1√1−(12)2 [∵cos−1x=sin−1√1−x2]=94sin−1√1−19=94sin−12√23=RHS
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