For 8 = 30°, verify that: sin20 = 2 sino cose
Answers
Answer:
Given θ=30
∘
To verify,
(i) tan2θ=
1−tan
2
θ
2tanθ
tan2(30
∘
)=
1−tan
2
30
∘
2tan30
∘
tan(60
∘
)=
1−tan
2
30
∘
2tan30
∘
3
=
1−(
3
1
)
2
2×
3
1
3
=
1−(
3
1
)
3
2
3
=
3
2
3
2
3
=
3
(ii) tan2θ=
1+tan
2
θ
4tanθ
We have, 1+tan
2
θ=sec
2
θ
∴tan2θ=
sec
2
θ
4tanθ
tan2(30
∘
)=
sec
2
30
∘
4tan30
∘
tan(60
∘
)=
sec
2
30
∘
4tan30
∘
3
=
(
3
2
)
2
4
3
1
3
=
3
4
3
4
3
=
3
3
3
=
3
(iii) cos2θ=
1+tan
2
θ
1−tan
2
θ
cos2(30
∘
)=
1+tan
2
30
∘
1−tan
2
30
∘
cos60
∘
=
1+tan
2
30
∘
1−tan
2
30
∘
2
1
=
1+(
3
1
)
2
1−(
3
1
)
2
2
1
=
3+1
3−1
2
1
=
4
2
2
1
=
2
1
(iv) cos3θ=4cos
3
θ−3cosθ
cos3(30
∘
)=4cos
3
30
∘
−3cos30
∘
cos(90
∘
)=4(cos
3
30
∘
)−3cos30
∘
0=4(
2
3
)
3
−3(
2
3
)
0=4(
8
3
3
)−3(
2
3
)
0=3(
2
3
)−3(
2
3
)
0=0