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C3- Trig help

How do you work out 6ii)??
Where does the pi/4 come from?

https://mei.org.uk/files/papers/C3_2015_June.pdf
Reply 1
Avatar for mqb2766
mqb2766
21
When is sin equal to cos?
Use cos(x) = sin(pi/2-x), hence its name.
Original post by oliviaaa98
How do you work out 6ii)??
Where does the pi/4 come from?

https://mei.org.uk/files/papers/C3_2015_June.pdf
Reply 2
Avatar for oliviaaa98
oliviaaa98
OP
9
does anyone have another way of explaining this? :confused:
Original post by oliviaaa98
How do you work out 6ii)??
Where does the pi/4 come from?

https://mei.org.uk/files/papers/C3_2015_June.pdf
Reply 3
Avatar for mqb2766
mqb2766
21
from the question you have
sin(x) = cos(x)
by defn this equals
sin(pi/2-x)
Hence
x = pi/2-x
...


Original post by oliviaaa98
does anyone have another way of explaining this? :confused:
Original post by oliviaaa98
does anyone have another way of explaining this? :confused:


Let x=sin⁡yx = \sin y. Due to the identity mentioned above, sin⁡x≡cos⁡(π2−x)\sin x \equiv \cos(\frac{\pi}{2}-x) it is true that x=cos⁡(π2−y)x = \cos(\frac{\pi}{2} - y).

Now take inverse of both sides...

arcsin⁡x=y\arcsin x = y
and
arccos⁡x=π2−y\arccos x = \frac{\pi}{2} - y.


If arcsin⁡x=arccos⁡x\arcsin x = \arccos x then it's true that y=π2−yy = \dfrac{\pi}{2} - y. Solve for yy. Since you are interested in what xx is, sub it back into x=sin⁡yx = \sin y.



An alternative approach (easier one) would be to first of all know how arcsin⁡x\arcsin x and arccos⁡x\arccos x look like plotted on a graph. Do that and I can explain this if you want.
(edited 4 years ago)
Reply 5
Avatar for oliviaaa98
oliviaaa98
OP
9
can you explain the graph method please, I've drawn out what they look like
Original post by RDKGames
Let x=sin⁡yx = \sin y. Due to the identity mentioned above, sin⁡x≡cos⁡(π2−x)\sin x \equiv \cos(\frac{\pi}{2}-x) it is true that x=cos⁡(π2−y)x = \cos(\frac{\pi}{2} - y).

Now take inverse of both sides...

arcsin⁡x=y\arcsin x = y
and
arccos⁡x=π2−y\arccos x = \frac{\pi}{2} - y.


If arcsin⁡x=arccos⁡x\arcsin x = \arccos x then it's true that y=π2−yy = \dfrac{\pi}{2} - y. Solve for yy. Since you are interested in what xx is, sub it back into x=sin⁡yx = \sin y.



An alternative approach (easier one) would be to first of all know how arcsin⁡x\arcsin x and arccos⁡x\arccos x look like plotted on a graph. Do that and I can explain this if you want.
Original post by oliviaaa98
can you explain the graph method please, I've drawn out what they look like


OK, so firstly reflect arcsin⁡x\arcsin x in the y-axis. What is the equation of the new curve? How does it look like?

Secondly, concerning this new curve, translate it vertically up by π2\dfrac{\pi}{2}. What is the equation of *this* curve? How does it look like?

Notice anything between this newest curve and arccos⁡x\arccos x?
Reply 7
Avatar for oliviaaa98
oliviaaa98
OP
9
so would the equation of the new curve be y=arcsin(-x)
and when you translate it the new equation is y=arcsin(-x) + pi/2
and it looks like a translation (-1,0) in the x direction of arccosx
??
Original post by RDKGames
OK, so firstly reflect arcsin⁡x\arcsin x in the y-axis. What is the equation of the new curve? How does it look like?

Secondly, concerning this new curve, translate it vertically up by π2\dfrac{\pi}{2}. What is the equation of *this* curve? How does it look like?

Notice anything between this newest curve and arccos⁡x\arccos x?
Original post by oliviaaa98
so would the equation of the new curve be y=arcsin(-x)
and when you translate it the new equation is y=arcsin(-x) + pi/2
and it looks like a translation (-1,0) in the x direction of arccosx
??


Sorry, I meant to say you reflected it in the x-axis. And so the equation is y=−arcsin⁡xy=-\arcsin x. Luckily, this is in fact the exact same as as y=arcsin⁡(−x)y = \arcsin(-x). So if you drew that one, it's fine.

Proceeding to translation up, you should have y=−arcsin⁡x+π2y = -\arcsin x + \dfrac{\pi}{2} now.

However, your observation is incorrect. It is not a translation in the vector (-1,0) of arccos(x). Feel free to use https://www.desmos.com/calculator/trffne7nsr to see what the final curve should look like. Note that the red curve is arccos(x).

Let's see if you notice anything between arccos⁡x\arccos x and this new curve now.
Reply 9
Avatar for oliviaaa98
oliviaaa98
OP
9
so its the same as arccosx?
Original post by RDKGames
Sorry, I meant to say you reflected it in the x-axis. And so the equation is y=−arcsin⁡xy=-\arcsin x. Luckily, this is in fact the exact same as as y=arcsin⁡(−x)y = \arcsin(-x). So if you drew that one, it's fine.

Proceeding to translation up, you should have y=−arcsin⁡x+π2y = -\arcsin x + \dfrac{\pi}{2} now.

However, your observation is incorrect. It is not a translation in the vector (-1,0) of arccos(x). Feel free to use https://www.desmos.com/calculator/trffne7nsr to see what the final curve should look like. Note that the red curve is arccos(x).

Let's see if you notice anything between arccos⁡x\arccos x and this new curve now.
Original post by oliviaaa98
so its the same as arccosx?


Yes, so we have an identity arccos⁡x≡π2−arcsin⁡x\arccos x \equiv \dfrac{\pi}{2} - \arcsin x.

You can use it in your question.
Reply 11
Avatar for oliviaaa98
oliviaaa98
OP
9
ok thanks!
Original post by RDKGames
Yes, so we have an identity arccos⁡x≡π2−arcsin⁡x\arccos x \equiv \dfrac{\pi}{2} - \arcsin x.

You can use it in your question.

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