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May 31st, 2012, 12:16 PM   #1
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trigonometry practice paper

Hello all!

I always find trigonometric formulas awkward, and I'm stuck on a practice paper; any help?

The acute angle (theta) is such that tan (theta) = 3 / 2

a) By considering an appropriate right-angled triangle, find the exact values of sin (theta) and cos (theta)

is this simply sin (theta) = 3 and cos (theta) = 2 ? or is it sin (theta) = + or - 15/4 cos (theta) = + or - 13/4 or something else??

I tried to follow some formulas; don't know if I did them right cos (theta) = + or - 1 / square root 1 + tan^2 theta and sin (theta) = + or - tan (theta) / square root 1 + tan^2 theta

b) Use the values in part (a), and trigonometric formulas, to find the exact values of cos (2 theta) and sin (2 theta)

well I don't have a clue, I'm looking at different double-angle formulas and all sorts, completely lost!

They are only given two marks each question, so I assuming they're not that complex;
an explanation would be handy, thanks again!
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May 31st, 2012, 12:22 PM   #2
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Re: trigonometry practice paper

a) Set up a right triangle such that the side opposite ? is 3 and the side adjacent is 2. What is the hypotenuse?

b) What are the double-angle identities for sine and cosine?
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May 31st, 2012, 01:25 PM   #3
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Re: trigonometry practice paper

ok

for a) hypotenuse is square root 13

so sin = 3 / square root 13

and cos = 2 / square root 13

for b) I used the identity

sin (2 theta) = 2 sin(theta)cos(theta) so,

sin (2 theta) = 2 sin (3 / square root 13) (cos (2 / square root 13) could I put this as 2 ( answer ) or leave it open such as 2 sin answer cos answer?

then for cos (2 theta) I used = 2 cos ^2 - 1

cos (2 theta) = 2 (cos (2 / square root 13)^2 - 1

Is this right?
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May 31st, 2012, 01:34 PM   #4
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You should simplify your answers by, for example, replacing the square of the square root of 13 with 13. What are your simplified results?
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May 31st, 2012, 02:08 PM   #5
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Re: trigonometry practice paper

Hey skipjack

do you mean, instead of

cos (2 theta) = 2 (cos (2 / square root 13)^2) - 1

change to

cos (2 theta) = 2 (cos ( 4 / 13 )) - 1

also could other possible simplifying involve, say

sin = 3 / square root 13 change to sin = 3 (square root 13) / 13 ? surds?
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May 31st, 2012, 11:48 PM   #6
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Try to be more precise: cos ? = 2/?13, so cos 2? = 2cos? - 1 = 2(4/13) - 1 = -5/13.

Unless you've been taught otherwise, it isn't necessary to write 2/?13 as 2?13 / 13.

Using the identity sin 2? ? 2 sin ? cos ?, what simple fraction do you now get for sin 2??
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June 1st, 2012, 06:14 AM   #7
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Re: trigonometry practice paper

so would sin 2? ? 2 sin ? cos ? be

sin 2? = 2 sin ( 3 / square root 13 ) cos ( 2 / square root 13 )

sin 2? = 2 sin cos ( 6 / square root 13 )

sin 2? = sin cos ( 12 / square root 13 )
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June 1st, 2012, 07:35 AM   #8
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Re: trigonometry practice paper

Not quite. We have so we replace the trigonometric functions with their values:

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June 1st, 2012, 09:26 AM   #9
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Re: trigonometry practice paper

Ah I see, I don't know why I thought that theta should only be replaced by the values found earlier (like a variable in its own right?!),

but we proved that sin (theta) and cos (theta) gave certain values already, so yeah it makes sense just to replace the trig function with the value we obtained like you said!

Thanks again MarkFL and skipjack for the help and not giving up on me!
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