Further Maths Trig Question

Angle Y is acute.

where p is a constant greater than 1.

Work out the expression for sin y in the form (attached picture)

Angle Y is acute.

where p is a constant greater than 1.

Work out the expression for sin y in the form (attached picture)

Others got to it before me haha.
I don't think this is a further maths problem though. Maybe C3

Sketch out a right angle triangle using the info of tan(y)
Then work out an expression for the hypotenuse.
Write down sin(y) (sin=O/H)
expand and simplify denominator


you should get sin(y) =p+1 / square root (2p+2)

Draw a triangle since we know the opposite = p + 1 and adjacent = p - 1

Do Pythagoras and work out the hypotenuse.

Then Sin(y) = opposite/hypotenuse

Draw a right-angled triangle, mark one angle as $y$, and the side opposite of it as $p+1$ while the side adjacent to it as $p-1$ and move on from there to find $\sin(y)$

Thanks guys, it was a lot easier than I thought it was.
The part I was missing was the right angle triangle

Extension exercise if you feel like it:

Try the question again but this time angle Y is reflex instead of acute.

I have no idea how you would do this...
Would you like get 360 minus Y?
Or would you use the trigonometric graph and find the corresponding value in the area between 90 and 180?

I have no idea how you would do this...
Would you like get 360 minus Y?
Or would you use the trigonometric graph and find the corresponding value in the area between 90 and 180?

Ignoring signs (plus/minus), the right-angle triangle relationships between sin/cos/tan are true for any angle. And you can see this using the identities e.g. for sin/cos you have

$\cos \theta = \pm \sqrt{1-\sin^2 \theta}$

You need to think about whether the final answer should be positive / negative and then whether the fraction needs to have a sign added.

And because between 90 and 180 sin X is positive, it must be a plus sign?
(Sorry late reply was having lunch)

And because between 90 and 180 sin X is positive, it must be a plus sign?
(Sorry late reply was having lunch)

Also I said the angle was reflex which is an angle greater than 180

That's not right and I'm no sure how you got it.

What answer did you get for the original question?

Also I said the angle was reflex which is an angle greater than 180

That's not right and I'm no sure how you got it.

What answer did you get for the original question?

Also I said the angle was reflex which is an angle greater than 180