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)

9.9.2T-Ratios of some special angles

i) Sin22 ½

Solution: sin2A = (1-cos2A)/2

Sin² 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: cos²A = (1 + cos2A)/2

cos²22 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)

$t ratios of multiple and sub multiple angles$

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

$t ratios of multiple and sub multiple angles$

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