Not sure if i've got the trig functions correct

[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)

Please help me

 

[groom03]

just looked and i don't think i should have divided by sin^2(x) because the powers on sin/cos would be wrong... i think

 

[snipez90]

just looked and i don't think i should have divided by sin^2(x) because the powers on sin/cos would be wrong... i think

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

 

[Redbelly98]

(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]

I'm doing A2 Edexcel maths and i keep on forgetting the trig functions …

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!

 

[groom03]

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 realized the second equation should equal cosec^2(x).

Thanks for your help

 

[Integral]

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:

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