# Not sure if i've got the trig functions correct

1. Sep 7, 2008

### groom03

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)

2. Sep 7, 2008

### groom03

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

3. Sep 7, 2008

### snipez90

$$\frac{a^2}{b^2} = (\frac{a}{b})^2, b \neq 0$$

4. Sep 7, 2008

### Redbelly98

Staff Emeritus
tan = sin/cos.

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

5. Sep 7, 2008

### tiny-tim

Hi groom03!

hmm … how to remember trigonometric identities … ?

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 and think cosec2 + cot2 = 1, then I multiply by sin2 and get 1 + cos2 = sin2 … which it isn't!

6. Sep 8, 2008

### groom03

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

$$\sin^2(x) + \cos^2(x) =1$$