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secondary 3 | A Maths
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Date Posted: 3 years ago
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We can make use of the following identity,
sin(A - B) = sin(A)cos(B) - sin(B)cos(A). [1]
The question is equivalent to,
sqrt{3}cos(x) - sin(x) = 1 [2]
We know that sqrt{3}/2 = sin(60) and cos(60) = 1/2. Substituting that in [2] yields,
2sin(60)cos(x) - 2sin(x)cos(60) = 1
Using the identity in [1], the LHS reduces to
2sin(60 - x) = 1
Which follows that,
sin(60° - x) =1/2
By taking the sine inverse on both sides, the basic angle is obtained.
Basic angle = 30°
60° - x = 30°
Which follows that x = 30°
You could account for the periodicity of sine, however you will realise that this would earn you values that are outside of the given restrictions.
sin(A - B) = sin(A)cos(B) - sin(B)cos(A). [1]
The question is equivalent to,
sqrt{3}cos(x) - sin(x) = 1 [2]
We know that sqrt{3}/2 = sin(60) and cos(60) = 1/2. Substituting that in [2] yields,
2sin(60)cos(x) - 2sin(x)cos(60) = 1
Using the identity in [1], the LHS reduces to
2sin(60 - x) = 1
Which follows that,
sin(60° - x) =1/2
By taking the sine inverse on both sides, the basic angle is obtained.
Basic angle = 30°
60° - x = 30°
Which follows that x = 30°
You could account for the periodicity of sine, however you will realise that this would earn you values that are outside of the given restrictions.
Hi, can u do it on paper? Then k can look at the comments also and understand better.
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