मराठी

Trigonometric Functions of Angles of a Triangle

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Estimated time: 4 minutes
Maharashtra State Board: Class 12

Formula: Trigonometric Functions of Angles of a Triangle

i. If A, B, and C are angles of a triangle ABC, then
A + B + C = π

a. sin (B + C) = sin (π − A) = sin A
sin (C + A) = sin B
sin (A + B) = sin C

b. cos (B + C) = cos (π − A) = − cos A
cos (C + A) = − cos B
cos (A + B) = − cos C

c. tan (B + C) = tan(π − A) = − tan A
tan (C + A) = − tan B
tan (A + B) = − tan C

ii. If A + B + C = π, then \[\frac{\mathrm{A+B}}{2}=\frac{\pi}{2}-\frac{\mathrm{C}}{2},\] \[\frac{\mathrm{C+A}}{2}=\frac{\pi}{2}-\frac{\mathrm{B}}{2}\mathrm{and}\frac{\mathrm{B+C}}{2}=\frac{\pi}{2}-\frac{\mathrm{A}}{2}\]

a. \[\sin\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)=\sin\left(\frac{\pi}{2}-\frac{\mathrm{C}}{2}\right)=\cos\frac{\mathrm{C}}{2}\]

\[\sin\left(\frac{\mathrm{B+C}}{2}\right)=\cos\frac{\mathrm{A}}{2}\]

\[\sin\left(\frac{\mathrm{C+A}}{2}\right)=\cos\frac{\mathrm{B}}{2}\]

b. \[\cos\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)=\sin\frac{\mathrm{C}}{2}\]

\[\cos\left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\sin\frac{\mathrm{A}}{2}\]

\[\cos\left(\frac{\mathrm{C}+\mathrm{A}}{2}\right)=\sin\frac{\mathrm{B}}{2}\]

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