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Trigonometric Identities Question Watch

1. Hi guys, I've been doing C2 past papers as part of revision and I've been through a fair few. I've been alright with these types of questions before but I can't get my head around this particular question, I have no idea why. This question is straight from a past paper - January 2010 (on the AQA board).

I've looked at the mark scheme and it doesn't help as it only gives the end answer, while the 'worked solutions' provided by my teacher is only an extended version of the answer and doesn't help towards reaching it.

8a) Solve the equation values of in the interval

I tried re-arranging the equation to and found the inverse tan of that to but didn't reach a correct solution.

2. Well a starting point would be

x + 52 = 22

Followed by

x + 52 = 22 + 180

Followed by ....

Using the period of the tangent function
3. You can't rearrange tan like that, it's a fancy function rather than a factor taken outside - for a simple example look at (x+52)^2=(22)^2, you can't rearrange that to be (x)^2=(22)^2 - (52)^2.

As for the answer to this question, look at the periodic nature of the tan function, what values can y take such that tan(y)=tan(22)? Then replace y by x+52 and solve for x. The easy answer is y=22, so x+52=22, and x=-30. However, this is outside the range of 0 to 360, so try bigger values of y.
4. (Original post by TenOfThem)
Well a starting point would be

x + 52 = 22

Followed by

x + 52 = 22 + 180

Followed by ....

Using the period of the tangent function

(Original post by Hopple)
You can't rearrange tan like that, it's a fancy function rather than a factor taken outside - for a simple example look at (x+52)^2=(22)^2, you can't rearrange that to be (x)^2=(22)^2 - (52)^2.

As for the answer to this question, look at the periodic nature of the tan function, what values can y take such that tan(y)=tan(22)? Then replace y by x+52 and solve for x. The easy answer is y=22, so x+52=22, and x=-30. However, this is outside the range of 0 to 360, so try bigger values of y.
Oh right! So I'd just have to use the recurring tan graph with and add 180 to get the two figures inside the range, seems like I was just over-complicating it, thanks for your help!
5. Just to add, if you wish to expand tan(A+B), there is an identity involving this which goes:

tan(A+B)= (tan A +tan B)/(1- tanA* tanB)

Peace.

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Updated: April 13, 2013
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