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Find d^5/dx^5 g(x) for g(x)=sinx+5x^4
Prove the following:
[1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)]
3 answers
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Anonymous
938 views
Hello! Can someone please check and see if I did this right? Thanks! :)
Directions: Find the exact solutions of the equation in
1 answer
asked by
Maggie
830 views
Mathematics - Trigonometric Identities - Reiny, Friday, November 9, 2007 at 10:30pm
(sinx - 1 -cos^2x) (sinx + 1 - cos^2x) should
3 answers
asked by
Anonymous
778 views
Simplify sin x cos^2x-sinx
Here's my book's explanation which I don't totally follow sin x cos^2x-sinx=sinx(cos^2x-1)
1 answer
asked by
Tara
1,051 views
Q: If y=sinx/(1+tanx), find value of x not greater than pi, corresponding to maxima or minima value of y. I have proceeded thus-
5 answers
asked by
MS
998 views
Find the derivatives of the following
18. y=〖cos〗^4 x^4 ANSWER: tanx2 19. y= sinx/(1+ 〖cos〗^2 x) ANSWER: 3sinx 20.
3 answers
asked by
StevE
649 views
y= 2sin^2 x
y=1- sinx find values of x inthe interval 0<x<360 if 2sin^x = 1-sinx this can be arranged into the quadratic. 2sin^2
0 answers
asked by
Candy
647 views
I need help solving for all solutions for this problem:
cos 2x+ sin x= 0 I substituted cos 2x for cos^2x-sin^2x So it became
1 answer
asked by
Martha
631 views
tanx+secx=2cosx
(sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 =
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asked by
shan
978 views
the problem is
2cos^2x + sinx-1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is
3 answers
asked by
alex
861 views