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secondary 3 | A Maths
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need help with this qn, pls explain too :)
Date Posted: 3 years ago
Views: 633
This formula is just a combination of a sine and a cosine into a single sine or cosine, by usage of the addition angle formulae.
You only need to remember the final forms
a sin @ + b cos @ = R sin (@ + taninv b/a)
a sin @ - b cos @ = R sin (@ - taninv b/a)
a cos @ + b sin @ = R cos (@ - taninv b/a)
a cos @ - b sin @ = R cos (@ + taninv b/a)
where R = sqrt (a^2 + b^2).
Basically, to transform the expression into a single sine, we put the sine first then the cosine, and likewise to transform the expression into a single cosine, we put the cosine first then the sine. This order is important as it affects the taninv b/a later on.
Pay attention to the signages.
You only need to remember the final forms
a sin @ + b cos @ = R sin (@ + taninv b/a)
a sin @ - b cos @ = R sin (@ - taninv b/a)
a cos @ + b sin @ = R cos (@ - taninv b/a)
a cos @ - b sin @ = R cos (@ + taninv b/a)
where R = sqrt (a^2 + b^2).
Basically, to transform the expression into a single sine, we put the sine first then the cosine, and likewise to transform the expression into a single cosine, we put the cosine first then the sine. This order is important as it affects the taninv b/a later on.
Pay attention to the signages.
thx :) can you help me with the practice now qns? i dont understand how the questions are solved even after looking at the worked example as theres no explanation and the workings are messy
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done
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We can just use the results of the formula. Unless your teacher really wants you to derive the full formula from scratch as a working.
Date Posted:
3 years ago
thx for helping me with this qn :)
done
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Q2
Note that if the cosine and the sine are arranged in the wrong way, we must swap their positions first.
For the two questions, there is no need to do so as the positions are already correct (furthermore, this is your first exercise for the R formula conversion so they shouldn’t possibly give you complicated ones as a start).
Note that if the cosine and the sine are arranged in the wrong way, we must swap their positions first.
For the two questions, there is no need to do so as the positions are already correct (furthermore, this is your first exercise for the R formula conversion so they shouldn’t possibly give you complicated ones as a start).
Date Posted:
3 years ago
i dont really understand the part where -7 is being factorised
For this, you must know the four forms of the R formula
The 5 cos x - 7 sin x takes the form
a cos x - b sin x = R cos (x + alpha)
where a and b are positive numbers and R = sqrt (a^2 + b^2)
The 5 cos x - 7 sin x takes the form
a cos x - b sin x = R cos (x + alpha)
where a and b are positive numbers and R = sqrt (a^2 + b^2)
the formula is Rsin(x+a), which has a positive sign but 5cosx-7sinx has a negative sign
since 5cosx-7sinx *negative* = Rsin(x+a) *positive* does that mean the negative sign is removed in 5cosx-7sinx?
sry i dont really know how to phrase it
since 5cosx-7sinx *negative* = Rsin(x+a) *positive* does that mean the negative sign is removed in 5cosx-7sinx?
sry i dont really know how to phrase it
In all the four formats
a sin x + b cos x = R sin (x + alpha)
a sin x - b cos x = R sin (x - alpha)
a cos x + b sin x = R cos (x - alpha)
a cos x - b sin x = R cos (x + alpha)
the numbers a and b are taken to be positive.
We just identify whether the signages are positive or negative before determining the form to take, but the values are still counted as positive.
So for example
2 sin x + 3 cos x
contains a positive in front of both the 2 sin x and the 3 cos x, so the correct form to take is R sin (x + alpha) where a = 2 and b = 3.
If we wish to take on the cosine form, we rearrange the above as 3 cos x + 2 sin x before realising that the new form to take is R cos (x - alpha) where a = 2 and b = 3.
Another example is
3 cos x - 5 sin x
where we identify a positive in front of 3 cos x but a negative in front of the 5 sin x, so the correct form to take is R cos (x + alpha) where a = 3 and b = 5.
Do not attempt to put b as -5 as this leads to confusion later on.
If our expression contains a negative in front of the first term, such as
- 5 sin x + 3 cos x
and we wish to use the sine form, then we must extract out the negative sign out of the expression to get - (5 sin x - 3 cos x)
first before taking the correct form R sin (x - alpha) on the bracketed term where a = 5 and b = 3, before placing the overall negative sign in front after that.
Remember that a and b are always taken to be positive; the signages attached to them will determine the form to take on.
a sin x + b cos x = R sin (x + alpha)
a sin x - b cos x = R sin (x - alpha)
a cos x + b sin x = R cos (x - alpha)
a cos x - b sin x = R cos (x + alpha)
the numbers a and b are taken to be positive.
We just identify whether the signages are positive or negative before determining the form to take, but the values are still counted as positive.
So for example
2 sin x + 3 cos x
contains a positive in front of both the 2 sin x and the 3 cos x, so the correct form to take is R sin (x + alpha) where a = 2 and b = 3.
If we wish to take on the cosine form, we rearrange the above as 3 cos x + 2 sin x before realising that the new form to take is R cos (x - alpha) where a = 2 and b = 3.
Another example is
3 cos x - 5 sin x
where we identify a positive in front of 3 cos x but a negative in front of the 5 sin x, so the correct form to take is R cos (x + alpha) where a = 3 and b = 5.
Do not attempt to put b as -5 as this leads to confusion later on.
If our expression contains a negative in front of the first term, such as
- 5 sin x + 3 cos x
and we wish to use the sine form, then we must extract out the negative sign out of the expression to get - (5 sin x - 3 cos x)
first before taking the correct form R sin (x - alpha) on the bracketed term where a = 5 and b = 3, before placing the overall negative sign in front after that.
Remember that a and b are always taken to be positive; the signages attached to them will determine the form to take on.
sorry for commenting here out of a sudden but for the cosine form at the first example, since it it rearranged to 3cosx + 2sinx, isnt it a=3 and b=2 or the a and b will always follow the main equation and not the rearranged one?
sorry for commenting here out of a sudden but for the cosine form at the first example, since it it rearranged to 3cosx + 2sinx, isnt it a=3 and b=2 or the a and b will always follow the main equation and not the rearranged one?
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Yes, I wrote wrongly and had not paid attention.
If y = 3 cos x + 2 sin x and we wish to transform this into a cosine form, then a = 3 and b = 2.
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Yes, I wrote wrongly and had not paid attention.
If y = 3 cos x + 2 sin x and we wish to transform this into a cosine form, then a = 3 and b = 2.
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