MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Composite Function

If A + B + C ...

Question

If $A + B + C = π$ , then prove that $s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2} - 1 = 4 s i n \frac{π - 4}{4} . s i n \frac{π - B}{4} . s i n \frac{π - C}{4}$

Open in App

Solution

$A + B + C = π \Rightarrow A + B = π - C (1) B + C = π - A (2) A + C = π - B (3)$
To Prove: $s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2} - 1 = 4 s i n \frac{π - 4}{4} . s i n \frac{π - B}{4} . s i n \frac{π - C}{4} L H S = s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2} - 1 = s i n \frac{A}{2} + s i n \frac{B}{2} + s i n \frac{C}{2} - s i n \frac{π}{2} (∵ 1 = s i n \frac{π}{2}) = 2 s i n \frac{\frac{A}{2} + \frac{B}{2}}{2} c o s \frac{\frac{A}{2} - \frac{B}{2}}{2} + 2 c o s \frac{\frac{C}{2} + \frac{π}{2}}{2} s i n \frac{\frac{C}{2} - \frac{π}{2}}{2} = 2 s i n \frac{A + B}{4} c o s \frac{A - B}{4} + 2 c o s \frac{C + π}{4} s i n \frac{C - π}{4} = 2 s i n \frac{π - C}{4} c o s \frac{A - B}{4} + 2 c o s \frac{C + π}{4} (- s i n \frac{π - C}{4}) = 2 s i n \frac{π - C}{4} [c o s \frac{A - B}{4} - c o s \frac{C + π}{4}] = 2 s i n \frac{π - C}{4} [- 2 s i n \frac{\frac{A - B}{4} + \frac{C + π}{4}}{2} s i n \frac{A - B}{4} - \frac{C + π}{2}] = - 4 s i n \frac{π - C}{4} [s i n \frac{(A + C) - B + π}{8} s i n \frac{A + π - (B + C)}{8}] = - 4 s i n \frac{π - C}{4} [s i n \frac{(A + C) - B + π}{8} s i n \frac{A + π - (π - 4)}{8}]$
$= - 4 s i n \frac{π - C}{4} [s i n \frac{2 π - 2 B}{8} s i n \frac{2 A - 2 π}{8}] = - 4 s i n \frac{π - C}{4} [s i n \frac{π - B}{4} s i n \frac{A - π}{4}] = 4 s i n \frac{π - C}{4} s i n \frac{π - B}{4} 4 s i n \frac{π - A}{4} = R H S$
Hence Proved.

Suggest Corrections

thumbs-up

1

Similar questions

A + B + C = π

sin (\frac{A}{2}) + sin (\frac{B}{2}) + sin (\frac{C}{2})

1 + 4 sin \frac{π - A}{4} sin \frac{π - B}{4} sin \frac{π - C}{4}

Q. Find the value of:

\frac{\frac{π}{4} sin \frac{π}{4}}{sin \frac{π}{2} - sin \frac{π}{4}}

Q. Prove that
sin ²(π/8 + A/2) - sin² (π/8 - A/2) = sin π/4 * sin A

A + B + C = π

(sin A + sin B + sin C) (- sin A + sin B + sin C) (sin A - sin B + sin C) (sin A + sin B - sin C) = 4 {sin}^{2} A {sin}^{2} B {sin}^{2} C

If A+B+C=180degree. Prove that sin2A+sin2B+sin2C=4×sinA×sinB×sinC.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Composite Functions

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Composite Function

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon