Revision: 11th Std >> Trigonometry - II MAH-MHT CET (PCM/PCB) Trigonometry - II

Two angles are said to be allied when their sum or difference is either 0 (zero) or an integral multiple of n/2.

Examples:

• −θ,  π2 ± θ,  π ± θ, 3π2 ± θ, 2π ± θ,etc.

(i) sin(A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C − sin A sin B sin C

or

sin(A + B + C) = cos A cos B cos C (tan A + tan B + tan C − tan A tan B tan C)

(ii) cos(A + B + C) = cos A cos B cos C − sin A sin B cos C − sin A cos B sin C − cos A sin B sin C

or

cos(A + B + C) = cos A cos B cos C (1 − tan A tan B − tan B tan C − tan C tan A)

(iii) tan(A + B + C) = \[=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}\]

(iv) cot(A + B + C) \[=\frac{\cot A\cot B\cot C-\cot A-\cot B-\cot C}{\cot A\cot B+\cot B\cot C+\cot C\cot A-1}\]

Note:

• 1 + cos 2θ = 2 cos²θ
• 1 − cos 2θ = 2 sin²θ

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

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

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

L.H.S. = `(tan theta)/(sec theta - 1)`

= `(tan theta)/(sec theta - 1) xx (sec theta + 1)/(sec theta + 1)`

= `(tan theta  (sec theta + 1))/(sec^2 theta - 1)`  ...[a² - b² = (a + b)(a - b)]

= `(tan theta  (sec theta + 1))/(tan^2 theta)  ...[(1 + tan^2 theta = sec^2theta),(tan^2theta = sec^2theta - 1)]`

= `(cancel(tan theta)  (sec theta + 1))/(cancel(tan^2 theta)_(tan theta))`

= `(sec theta + 1)/(tan theta)`

L.H.S. = R.H.S.

Hence proved.

If (A + B + C = 180°):

1. sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C

2. cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C

3. cos 2A + cos 2B − cos 2C = 1 − 4 sin A sin B cos C

4. \[\sin A+\sin B+\sin C=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\]

5. \[\cos A+\cos B+\cos C=1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\]

6. cos A + cos B − cos C = \[-1+4\cos\frac{A}{2}\cos\frac{B}{2}\sin\frac{C}{2}\]

7. tan A + tan B + tan C = tan A tan B tan C

8. cot A cot B + cot B cot C + cot C cot A = 1

9. \[\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}=1\]

10. \[\cot\frac{A}{2}+\cot\frac{B}{2}+\cot\frac{C}{2}=\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}\]

sin(nπ + θ) = (−1)ⁿ sin θ

sin(nπ − θ) = (−1)ⁿ⁻¹ sin θ

cos(nπ ± θ) = (−1)ⁿ cos θ

\[\sin\frac{A}{2}\pm\cos\frac{A}{2}=\pm\sqrt{1\pm\sin A}\]

\[\frac{1-\cos\alpha}{\sin\alpha}=\tan\frac{\alpha}{2},\alpha\neq(2n+1)\pi\]

\[\frac{1+\cos\alpha}{\sin\alpha}=\cot\frac{\alpha}{2},\alpha\neq2n\pi\]

\[\frac{1-\cos\alpha}{1+\cos\alpha}=\tan^{2}\frac{\alpha}{2},\alpha\neq(2n+1)\pi\]