Fundamental trignometric identities?

Question

I'm having serious troubles with these and have been working on them for some time. I'm mathtarded and have a major headache over these.

tanx ( sinx + cotx cosx) = secx
sin x (csc x - sin x ) = cos^2x
cos^2x + tan^2x + cos^2 x = 1
sec^2x + tan^2x sec^2x = sec^4 x


can anyone help me through these? :(

The third one is typed correctly, I guess my math teacher messed up !! haha. Thanks for that guys, they were real had my stumped. I've gotten all the easier ones, but the hardest ones I just can't seem to do. Could I get help with one more?


(cosx - sinx)^2 = 1 - 2cosx sinx

Answer 1

Draw a right triangle. Label the right angle and hypotenuse as "C". label the smallest angle and opposite side as "X"; then the remaining as "B".

Then substitute the ratios of sides for the trig functions. Then cancel things out and multiply together.

Finally convert the ratios back to trig functions.

Answer 2

tanx(sinx + cotx cosx) = tanx (sinx +(cos x)^2/sin x)
=tanx(sin^2 x+cos^2 x)/sin x
=tanx/sinx = 1/cosx
=sec x

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

cos^2x + tan^2 x+cos^2 x is not equal to 1 (I assume you have typed it wrong)

sec^2x + tan^2x sec^2x = sec^2x (1 + tan^2x) = sec^2x * sec^2x = sec^4 x

Most of these just use the identity sin^2x+cos^2x = 1. A related identity was used in the last one, tan^2x + 1 = sec^2 x.

Answer 3

tanx ( sinx + cotx cosx) = secx
(sinx / cosx) (sinx + (cosx / sinx) cosx) = secx
(sinx / cosx) ((sin^2 x / sinx) + (cos^2 x / sinx)) = secx
(sinx / cosx) (1/sinx) = secx
1/cosx = secx
secx = secx

sin x (csc x - sin x ) = cos^2x
sinx (1/sinx - sinx) = cos^2 x
sinx (1/sinx - sin^2 x / sinx) = cos^2 x
sinx (cos^2 x / sinx) = cos^2 x
cos^2 x = cos^2 x

cos^2x + tan^2x + cos^2 x = 1
are you sure you typed this right?

sec^2x + tan^2x sec^2x = sec^4 x
sec^2 x (1 + tan^2 x) = sec^4 x
sec^2 x (sec^2 x) = sec^4 x
sec^4 x = sec^4 x

Answer 4

To make it easier, let me just use s for sin and c for cos, and I will convert all expressions to sin and cos, so tan=s/c; sec=1/c; csc = 1/s

a) s/c[s+(c/s)c]=[s^2/c + c]=[(s^2+c^2)/c]=1/c=sec x [note sin^2+cos^2=1]

b) s[1/s -s]=s[(1-s^2)/s]=1-s^2=cos^2 (again, since c^2+s^2=1 => c^2=1-s^2]

c) please check that you have typed this correctly; tanx >1 for angles greater than 45, so if you square that number you cannot get a sum of squares to be 1, please check again

d) 1/c^2 + (s^2/c^2)(1/c^2)=1/c^2[1+s^2/c^2]=
1/c^2[(c^2+s^2)/c^2]=1/c^2[1/c^2]=1/c^4=sec^4

Answer 5

tanx (sinx + cotx cosx)

(sinx/cosx)[sinx + (cosx/sinx)cosx]

sin^2x/cosx + sinx/cosx)(cosx/sinx)(cosx)

sin^2x/cosx + cosx

sin^2x/cosx + cos^2x/cosx

(sin^2x + cos^2x)/cosx

1/cosx

secx

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sin x (csc x - sin x )

sinx(1/sinx - sinx)

sinx/sinx - sin^2x

1 - sin^2x

cos^2x

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cos^2x + tan^2x + cos^2 x

2cos^2x + sin^2x/cos^2x

Are you sure you wrote this one down correctly?

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sec^2x + tan^2x sec^2x

sec^2x(1 + tan^2x)

sec^2x sec^2x

sec^4x