# How do I tackle questions like these?

Announcements
#1
0
4 years ago
#2
(Original post by zattyzatzat)
...
First one can be written in terms of tan(x), since tan is periodic.
Second can be written in terms of cot.
Third can be acquired from drawing a right-angled triangle satisfying .
0
4 years ago
#3
parts i and ii can be done just be knowing what the graph of tan(x) looks like:

tan(x) is periodic on pi so tan(x+pi) = tan(x), so tan(pi-x) = tan(-x) = -tan(x) = -k since tan(x) = sin(x)/cos(x) is an odd function.

tan(1/2 pi - x) = 1/tan(x) = 1/k

the last one is best done by thinking about the right-angled triangle that gives rise to tan(x) = k = opp/adj -> so say k /1 = k, so the hypotenuse is (k^2 + 1)^1/2 so sin(x) = opp/hyp = k/(k^2 + 1)^1/2
2
4 years ago
#4
(Original post by Ryanzmw)
...
Have you read the sticky to do with forum guidelines?

https://www.thestudentroom.co.uk/sho...80&postcount=2
1
#5
(Original post by Ryanzmw)
parts i and ii can be done just be knowing what the graph of tan(x) looks like:

tan(x) is periodic on pi so tan(x+pi) = tan(x), so tan(pi-x) = tan(-x) = -tan(x) = -k since tan(x) = sin(x)/cos(x) is an odd function.

tan(1/2 pi - x) = 1/tan(x) = 1/k

the last one is best done by thinking about the right-angled triangle that gives rise to tan(x) = k = opp/adj -> so say k /1 = k, so the hypotenuse is (k^2 + 1)^1/2 so sin(x) = opp/hyp = k/(k^2 + 1)^1/2
I think understand how to do the first two now thanks. But the last question still confuses. How did you refer the opposite side of a triangle as K?
0
4 years ago
#6
(Original post by RDKGames)
Have you read the sticky to do with forum guidelines?

https://www.thestudentroom.co.uk/sho...80&postcount=2
Ah sorry my bad. I'll refrain from posting full solutions in the future.
1
4 years ago
#7
(Original post by zattyzatzat)
I think understand how to do the first two now thanks. But the last question still confuses. How did you refer the opposite side of a triangle as K?
tan(x) is the ratio of two sides of a right-angled triangle, right? (Remember SOH CAH TOA) so we can construct a right-angled triangle which will certainly have opp/adj = K. Then since it's right-angled we can find the third side by Pythagoras' theorem and calculate sin or cos
0
#8
(Original post by Ryanzmw)
tan(x) is the ratio of two sides of a right-angled triangle, right? (Remember SOH CAH TOA) so we can construct a right-angled triangle which will certainly have opp/adj = K. Then since it's right-angled we can find the third side by Pythagoras' theorem and calculate sin or cos
Ah I see thanks!
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Have you done work experience at school/college?

Yes (155)
41.67%
Not yet, but I will soon (69)
18.55%
No (148)
39.78%