Let two simple harmonic motions are y=a‌sin‌omega‌t and y=a‌sin(ωt+ϕ) In the first case,
a
2
=a‌sin‌ω‌t⇒sin‌ω‌t=
1
2
∴ cos‌ω‌t=
√3
2
In the second case,
a
2
=a‌sin(ωt+ϕ) ⇒
1
2
=[sin‌ω.cos‌ϕ+cos‌ω‌t‌sin‌ϕ] ⇒
1
2
=[
1
2
‌cos‌ϕ+
√3
2
‌sin‌ϕ] ⇒ 1−cos‌ϕ=√3‌sin‌ϕ ⇒ (1−cos‌ϕ)2=3sin2ϕ ⇒ (1−cos‌ϕ)2=3(1−cos2ϕ) On solving we get cos‌ϕ=+1‌or‌cos‌ϕ=