pleeeeeeeeeeeeeeeeeeeese help

[yasaminG]

in the expression 1/squareroot 16 - t^2, make the substitution t= 4 sin u and simplify result. answer is between 0<u<pi/2



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show each is an identity:

Sin A/ 1+Cos A + 1+Cos A / Sin A = 2 Csc A

and finally -

Sin A Cos A = tan A / 1+ tan^2 A


i appreciate which ever ones you can solve , thanks

 

[tkhunny]

in the expression 1/squareroot 16 - t^2, make the substitution t= 4 sin u and simplify result. answer is between 0<u<pi/2

That makes no sense. There isn't an "answer" without a problem statement. I'm sort of looking for an equals sign.

 

[yasaminG]

in the expression 1/squareroot 16 - t^2, make the substitution t= 4 sin u and simplify result. answer is between 0<u<pi/2

That makes no sense. There isn't an "answer" without a problem statement. I'm sort of looking for an equals sign.

thats exactly what the problem says in my book, the answer is 1/4 csc u if it helps

 

[tkhunny]

OK Make the substitution and simplfy. Have you forgotten your trig identities?

 

[yasaminG]

ok i really cant solve it, i get stuck with the sqaure rooting part of the sin

i get 1/4-2sinu

 

[Gene]

When you substitute t=4sin(u) you SHOULD get
1/sqrt(16-16sin²(u)) = Notice the ()s. They are ALL necessary
1/(4sqrt(1-sin²(u)))
Now if you can only remember that
sin²(u)+cos²(u)=1
you should be able to do something with it.

Sin(A)/(1+Cos(A))+(1+Cos(A))/Sin(A)=2Csc(A)
2Csc(A)=2/Sin(a)
Multiply by the common denominator
(1+Cos(A))Sin(A) and see what happens.

Sin(A)Cos(A) = tan(A)/(1+ tan^2(A)) Change tan(A) to Sin(A)/Cos(A) and use sin²(u)+cos²(u)=1 when appropriate.

 

[yasaminG]

thanks im trying to work out what youve written