# Math Assignment Help With T-Ratios Of Multiple And Sub Multiple Angles

## 9.9 T- Ratios of multiple and sub-multiple angles:

i)sin2A = 2sinAcosA

ii)cos2A = cos^{2}A – sin^{2}A = 1 – 2sin^{2}A = 2cos^{2}A – 1

iii)tan2A = 2tanA

1 – tan^{2}A

iv)1 – cos2A = 2sin^{2}A

v)1 + cos2A = 2cos^{2}A

vi)sin2A = 2tanA

1+ tan^{2}A

vii) cos2A = 1 - tan^{2}A

1+ tan^{2}A

viii) sin3A = (3sinA – 4sin^{3}A)

ix) cos3A = (4cos^{3}A – 3cosA)

x) tan3A = 3tanA – tan^{3}A

1 –3 tan^{2}A

**9.9.1 T-Ratios of sub-multiple angles**

Replacing A by A/2 in above results.

i) sinA = 2sin(A/2) cos(A/2)

ii) cosA = (cos^{2}(A/2) – sin^{2}(A/2))

iii) 1 – cosA = 2sin^{2}(A/2)

iv) 1 + cosA = 2cos^{2}(A/2)

v) tanA = 2tan(A/2)

1 – tan^{2}(A/2)

vi) sinA = 2tan(A/2)

1 + tan^{2}(A/2)

vii) cosA = 1 – tan^{2}(A/2)

1 + tan^{2}(A/2)

**9.9.2** **T-Ratios of some special angles**

**i)****Sin22 ½**

**Solution:** sin2A = (1-cos2A)/2

Sin^{2} 22 ½ = (1 – cos45^{o})/2

= (1 – 1/√2)/2

= (√2 – 1)/2√2

Sin 22 ½ = √(√2 – 1)/2√2

ii) **Cos22 ½ **

**Solution**: cos^{2}A = (1 + cos2A)/2

cos^{2}22 ½ = ( 1+ cos 45^{o})/2

= (1 + 1/√2)/2

= (√2 + 1)/2√2

cos 22 ½ =√ (√2 +1)/2√2

**iii) Tan22 ½ **

**Solution:** tan22 ½ = sin22 ½

cos22 ½

put the values of sin and cos as obtained above and solve.

Tan22 ½ = (√2 – 1)

v) **Tan11 ½**

**Solution:** tanA = 2 tan(A/2)

1 – tan^{2}(A/2)

Tan22 ½ = 2 tan11 ¼

1 – tan^{2}11 ¼

Take tan11 ¼ = *x*

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