Not sure if i've got the trig functions correct

  • Thread starter groom03
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27
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I'm doing A2 Edexcel maths and i keep on forgetting the trig functions so can someone take a look and tell me if i've got it right.

So far:

Sin^2(x)=cos^2(x)=1

so:

sin^2(x)=1-cos^2(x)

cos^2(x)=1-sin^2(x)

this is where i get abit stuck

(sin^2(x)+cos^2(x)=1)/cos^2(x) = tan^2(x)+1=sec^2(x)

(sin^2(x)+cos^2(x)=1)/sin^2(x) = tan^2(x)+1=sec^2(x)

i'm not sure if i've got them wrong or if i'm meant to divide by sin(x) or sin^2(x)

Please help me :smile:
 
27
0
just looked and i dont think i should have divided by sin^2(x) because the powers on sin/cos would be wrong.... i think
 
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3
just looked and i dont think i should have divided by sin^2(x) because the powers on sin/cos would be wrong.... i think
[tex]\frac{a^2}{b^2} = (\frac{a}{b})^2, b \neq 0[/tex]
 

Redbelly98

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(sin^2(x)+cos^2(x)=1)/cos^2(x) = tan^2(x)+1=sec^2(x)

(sin^2(x)+cos^2(x)=1)/sin^2(x) = tan^2(x)+1=sec^2(x)
tan = sin/cos.

1st equation here looks good.
2nd equation: you get a cos/sin term, that is not = tan.
 

tiny-tim

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I'm doing A2 Edexcel maths and i keep on forgetting the trig functions …
Hi groom03! :smile:

hmm … how to remember trigonometric identities … ? :rolleyes:

I always find that the safest plan is to write down what I think the formula is, and then multiply by cos2 or sin2 to check it.

For example, if I get confused :confused: and think cosec2 + cot2 = 1, then I multiply by sin2 and get 1 + cos2 = sin2 … which it isn't! :wink:
 
27
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tan = sin/cos.

1st equation here looks good.
2nd equation: you get a cos/sin term, that is not = tan.

i keep on getting sin/cos and cos/sin wrong, and i just realised the second equation should equal cosec^2(x).

Thanks for your help
 

Integral

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I'm doing A2 Edexcel maths and i keep on forgetting the trig functions so can someone take a look and tell me if i've got it right.

So far:

Sin^2(x)=cos^2(x)=1
I am guessing that this is a typo. You surely mean:

[tex] \sin^2(x) + \cos^2(x) =1 [/tex]
 

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