# 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 = cos2A – sin2A = 1 – 2sin2A = 2cos2A – 1

iii)tan2A = 2tanA

1 – tan2A

iv)1 – cos2A = 2sin2A

v)1 + cos2A = 2cos2A

vi)sin2A = 2tanA

1+ tan2A

vii) cos2A = 1 - tan2A

1+ tan2A

viii) sin3A = (3sinA – 4sin3A)

ix) cos3A = (4cos3A – 3cosA)

x) tan3A = 3tanA – tan3A

1 –3 tan2A

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 = (cos2(A/2) – sin2(A/2))

iii) 1 – cosA = 2sin2(A/2)

iv) 1 + cosA = 2cos2(A/2)

v) tanA = 2tan(A/2)

1 – tan2(A/2)

vi) sinA = 2tan(A/2)

1 + tan2(A/2)

vii) cosA = 1 – tan2(A/2)

1 + tan2(A/2)

9.9.2 T-Ratios of some special angles

1. i) Sin22 ½

Solution: sin2A = (1-cos2A)/2

Sin2 22 ½ = (1 – cos45o)/2

= (1 – 1/√2)/2

= (√2 – 1)/2√2

Sin 22 ½ = √(√2 – 1)/2√2

ii) Cos22 ½

Solution: cos2A = (1 + cos2A)/2

cos222 ½ = ( 1+ cos 45o)/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 – tan2(A/2)

Tan22 ½ = 2 tan11 ¼

1 – tan211 ¼

Take tan11 ¼ = x

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