Question
can i use factoring to simplify this trig identity?
the problem is sinx + cotx * cosx i know the answer is cscx and i know how to get it but i want to know if i can do factoring to get it bc i tried to but it wont give me the answer .
this is the step i went through:
1) sinx + cotx * cosx turns into sinx +(cosx/sinx)*cosx
2) i try to factor out sinx so that i would get sinx(1+ cosx * cosx)
3) that left me with sinx(1+cos^(2) x) that's where i im lost can anyone enlighten me plz
the problem is sinx + cotx * cosx i know the answer is cscx and i know how to get it but i want to know if i can do factoring to get it bc i tried to but it wont give me the answer .
this is the step i went through:
1) sinx + cotx * cosx turns into sinx +(cosx/sinx)*cosx
2) i try to factor out sinx so that i would get sinx(1+ cosx * cosx)
3) that left me with sinx(1+cos^(2) x) that's where i im lost can anyone enlighten me plz
Answers
cot = cos/sin, so you have
sin + cos^2/sin
= (sin^2+cos^2)/sin
= 1/sin
= csc
sin + cos^2/sin
= (sin^2+cos^2)/sin
= 1/sin
= csc
im well aware of that steve thank you for answering but i really wanted to know was is it at all possible to use factoring to solve this like i have up there
I don't see any way to use factoring. You don't in fact come up with
sin(1+cos^2)
because you have that pesky 1/sin under the cos^2.
If you try to fractor out the sin, you get
sin(1+cos^2/sin^2)
and again you end up with sin^2+cos^2 on top.
sin(1+cos^2)
because you have that pesky 1/sin under the cos^2.
If you try to fractor out the sin, you get
sin(1+cos^2/sin^2)
and again you end up with sin^2+cos^2 on top.
We could do some " silly" factoring
sinx + (cosx/sinx)(cosx)
= sinx + cos^2 x (sinx)^-1
= (sinx)^-1 (sin^2 x + cos^2 x)
= (sinx)^-1
= 1/sinx
= cscx
sinx + (cosx/sinx)(cosx)
= sinx + cos^2 x (sinx)^-1
= (sinx)^-1 (sin^2 x + cos^2 x)
= (sinx)^-1
= 1/sinx
= cscx
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