How do I tackle questions like these?

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zattyzatzat
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RDKGames
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(Original post by zattyzatzat)
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First one can be written in terms of tan(x), since tan is \pi periodic.
Second can be written in terms of cot.
Third can be acquired from drawing a right-angled triangle satisfying \tan x = k.
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Ryanzmw
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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
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RDKGames
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(Original post by Ryanzmw)
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Have you read the sticky to do with forum guidelines?

https://www.thestudentroom.co.uk/sho...80&postcount=2
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zattyzatzat
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(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?
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Ryanzmw
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(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.
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Ryanzmw
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(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
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zattyzatzat
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(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!
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