If ????=30°, verify that:
(i)tan2????=2tan????/1-tan²????
(ii)sin2????=2tan????/1+tan²????
(iii)cos2????=1-tan²????/1+tan²????
(iv)cos3????=4cos3????-3cos????
Answers
If θ = 30°, verify that:
(i)tan2θ = 2tanθ/1-tan²θ
(ii)sin2θ = 2tanθ/1+tan²θ
(iii)cos2θ = 1-tan²θ/1+tan²θ
(iv)cos3θ = 4cos³θ - 3cosθ
(i) LHS = tan2θ = tan2(30°) = tan60° = √3
RHS = 2tanθ/(1 - tan²θ) = 2tan30°/(1 - tan²30°)
= 2 × 1/√3/(1 - 1/√3²)
= 2/√3/(1 - 1/3)
= 2/√3/(2/3)
= √3
LHS = RHS hence verified
(ii) LHS = sin2θ = sin2(30°) = sin60° = √3/2
RHS = 2tanθ/(1 + tan²θ) = 2tan30°/(1 + tan²30°)
= 2 × 1/√3/(1 + 1/√3²)
= 2/√3/( 1 + 1/3)
= 2/√3/(4/3)
= √3/2
LHS = RHS hence verified
(iii) LHS = cos2θ = cos2(30°) = cos60° = 1/2
RHS = (1 - tan²θ)/(1 + tan²θ) = (1 - tan²30°)/(1 + tan²30°)
= (1 - 1/√3²)/(1 + 1/√3²)
= (1 - 1/3)/(1 + 1/3)
= (2/3)/(4/3)
= 1/2
LHS = RHS hence verified
(iv)LHS = cos3θ = cos3(30°) = cos90° = 0
RHS = 4cos³θ - 3cosθ = 4cos³30° - 3cos30°
= 4(√3/2)³ - 3(√3/2)
= 4 × 3√3/8 - 3√3/2
= 3√3/2 - 3√3/2
= 0
LHS = RHS hence verified
Answer:
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