Question

in triangle ABC show that sin square A by 2 + sin square b + c by 2 is equal to

Answer 1

hey!!

Nice question...

sin²A/2+sin²B+C/2=1

As you know that the sum of all angle of triangle is 180°

<A+<B+C=180°

B+C=180°-A

B+C/2=180/2-A/2 {if we dividing by 2 on both side }

sin²(B+C/2)=sin²(90°-A/2} {multiplying by sin²on both side,}

sin²(B+C)=cos²A/2 ........1)

now 1 substituting on given question

we get ..

sin²A/2+cos²A/2

=>1 ...like[ sin²A+cos²A=1]

hope it helps you

@Rajukumar111

Answer 2

Answer:

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Step-by-step explanation:

sin²A/2+sin²B+C/2=1

As you know that the sum of all angle of triangle is 180°

<A+<B+C=180°

B+C=180°-A

B+C/2=180/2-A/2 {if we dividing by 2 on both side }

sin²(B+C/2)=sin²(90°-A/2} {multiplying by sin²on both side,}

sin²(B+C)=cos²A/2 ........1)

now 1 substituting on given question

we get ..

sin²A/2+cos²A/2

=>1 ...like[ sin²A+cos²A=1]

hope it helps you