If θ = 30°, verify that
(i)
(ii)
(iii)
(iv)cos 3θ = 4 cos³ θ − 3 cos θ
Answers
SOLUTION :
Given : θ = 30°
verify (i) tan 2θ = 2tanθ / 1 - tan²θ
L.H.S = tan 2θ
On Substituting θ = 30° in LHS
= tan 2 × 30°
= tan 60°
= tan 60° = √3
L.H.S = √3
RHS = 2tanθ / 1 - tan²θ
On Substituting θ = 30° in RHS
2 tan 30° / 1 - (tan 30°)²
= 2× 1/√3 / 1 - (1/√3)²
= 2/√3 / 1 - 1/3
= (2/√3) / (3 - 1)/3
[ By taking L.C.M ]
= (2/√3) /( ⅔)
= 2/√3 × 3/2
= 3/√3
= 3×√3 / √3×√3
[By rationalising the denominator]
= 3√3 /3
R.H.S = √3
L.H.S = R.H.S
Hence, tan 2θ = 2tanθ / 1 - tan²θ
(ii) Given : θ = 30°,
verify : sin 2θ = 2 tanθ/1− tan²θ
L.H.S = sin 2θ
On Substituting θ = 30° in LHS
= sin 2× 30°
sin 2θ = sin 60° = √3/2
L.H.S = √3/2
R.H.S = 2 tanθ/1 + tan²θ
= 2 tan 30° /1+ tan² 30°
On Substituting θ = 30° in RHS
= 2 × 1/√3 / 1+ (1/√3)²
= 2× 1/√3 / 1 + (1/√3)²
= 2/√3 / 1 + 1/3
= (2/√3) / (3 + 1)/3
[ By taking L.C.M ]
= (2/√3) /(4/3)
= 2/√3 × (¾)
= 3/2√3
= 3×√3 / 2√3×√3
[By rationalising the denominator]
= 3√3 /2×3
R.H.S = √3 /2
L.H.S = R.H.S
Hence, sin 2θ = 2 tanθ / 1 + tan²θ
(iii) Given : θ = 30°,
verify : cos 2θ = 1- tan² θ/1+ tan²θ
L.H.S = cos 2θ
On Substituting θ = 30° in LHS
= cos 2× 30°
cos 2θ = cos 60° = 1/2
L.H.S = 1/2
R.H.S = 1- tan² θ/1+ tan²θ
On Substituting θ = 30° in RHS
= 1 - tan² 30° / 1+ tan² 30°
= 1 - (1/√3)² / 1 + (1/√3)²
= 1 - ⅓ / 1 + ⅓
= (3-1)/3 / (3+1)/3
[By taking L.C.M]
= ⅔ / 4/3
= ⅔ × ¾
= ½
L.H.S = R.H.S
Hence, cos 2θ =1- tan²θ/1+tan²θ
(iv) Given : θ = 30°,
verify: cos 3θ = 4 cos³θ - 3cosθ
LHS = cos 3θ
On Substituting θ = 30° in LHS
= cos 3 (30°)
= cos 90°
= 0
[cos 90° = 0]
LHS = 0
RHS = 4 cos³θ −3 cosθ
On Substituting θ = 30° in RHS
= 4cos³ 30° −3cos 30°
= 4(√3/2)³ −3× √3/2
= 4 × 3√3/8 - 3√3/2
= 3√3/2 − 3√3/2
= 0
RHS = 0
LHS = RHS.
Hence, cos 3θ = 4 cos³θ - 3 cosθ
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