trig identities! fun fun!

[struntz]

hello everyone i can't seem to figure these few out....

solve these using fundamental identites...

here is an example of one i did...
u can use either side to start solving so it matches the opposite side.

sin^2x(1 + cot^2x) = 1

sin^2+sin^2xcot^2x

sin^2x + sinx cos^2x/sin^2x <---sin's cross out

sin^2x + cos^2x = 1 <----so there u go



1. cotx + tanx = secx cscx

2. sin^2x + tan^2 x + cos^2x = sec^2x

thanks all, later!

[lloydie]

Well, as no one else has answered your question, I'll give it a go...

1) cotx + tanx = secx cscx

[I'm afraid I don't know what cscx is. Do you mean cosecx?]
[If you do, reply to the post ajd I'll show you how it's done]
[If it's not supposed to be cosecx and it is cscx, i'm afraid I ]
[don't know how to solve it! ]

2) sin^2x + tan^2 x + cos^2x = sec^2x

[Use the rule: sec^2 = 1 + tan^2x] So...

sin^2x + tan^2 x + cos^2x = 1 + tan^2x

[Take tan^2x from both sides] So...

sin^2x + cos^2x = 1 <----so there u go!!

If you need any more help on this just reply to the post.
------------------
Lloyd Morgan
UK

[struntz]

hey thanks for ur reply but i don't think thats right, see what ur trying to do is make the 2 sides == to eachother, such as


if you have sin^2x+tan^2x + cos^2x = sec^2x

u can either start on teh left side and make it equal the right side or vice versa...

so say u start with the left side, u only have
sin^2x + tan^2x + cos^2x to work with, u must make that some how equal the right side

and yes
csc == cosine
sin == sine
sec == secant
cos == cosecant
tan == tangent
cot == cotangent

[struntz]

i think i figured out the one u tried but i still can't figure out the top one, the answer to

sin^2x + tan^2x + cos^2x = sec^2x

since sin^2x + cos^2x = 1 u can put

1 + tan^2x = sec^2x

and since 1 + tan^2x == sec^2x, ur done!

[lloydie]

Ah, yes, I'm afraid I only realised that also about an hour after I posted it.
I should have known that anyway, because I got set a load of questions to do only last week on this exact topic for A-level maths.
About the other question:

cotx + tanx = secx cosecx

At last I can be some use, as we got given this exact question last week and I got it right! Here it is... copying right off my paper:

cotx + tanx = secx cosecx

[tanx = sinx / cosx] Therefore
[as cotx = 1 / tanx ... cotx = cosx / sinx] So...

(cosx / sinx) + (sinx / cosx) = secx cosecx

[Make a common denominator & sort it out] So...

(cos^2x + sin^2x) / cosx sinx = secx cosecx

[cos^2x + sin^2x = 1] So...

1 / cosx sinx = secx cosecx

(1 / cosx) (1 / sinx) = secx cosecx

[(1 / cosx = secx) & (1 / sinx = cosecx)] So...

secx cosecx = secx cosecx

-------------------

Is this OK now? I hope so.

Regards,
Lloyd

[struntz]

excellent job, works great