Math
 👍 0
 👎 0
 👁 395
Respond to this Question
Similar Questions

Precalculus
Prove the following identities. 1. 1+cosx/1cosx = secx + 1/secx 1 2. (tanx + cotx)^2=sec^2x csc^2x 3. cos(x+y) cos(xy)= cos^2x  sin^2y

Trig.......
I need to prove that the following is true. Thanks (2tanx /1tan^x)+(1/2cos^2x1)= (cosx+sinx)/(cosx  sinx) and thanks ........... check your typing. I tried 30º, the two sides are not equal, they differ by 1 oh , thank you Mr

Math help again
cos(3π/4+x) + sin (3π/4 x) = 0 = cos(3π/4)cosx + sin(3π/4)sinx + sin(3π/4)cosx  cos(3π/4)sinx = 1/sqrt2cosx + 1/sqrt2sinx + 1/sqrt2cosx  (1/sqrt2sinx) I canceled out 1/sqrt2cosx and 1/sqrt2cosx Now I have 1/sqrt sinx +

Trig Identities
Prove the following identities: 13. tan(x) + sec(x) = (cos(x)) / (1sin(x)) *Sorry for any confusing parenthesis.* My work: I simplified the left side to a. ((sinx) / (cosx)) + (1 / cosx) , then b. (sinx + 1) / cosx = (cos(x)) /

Math
Suppose f(x) = sin(pi*cosx) On any interval where the inverse function y = f –1(x) exists, the derivative of f –1(x) with respect to x is: a)1/(cos(pi*cosx)), where x and y are related by the equation (satisfy the equation)

trigonometry
how do i simplify (secx  cosx) / sinx? i tried splitting the numerator up so that i had (secx / sinx)  (cosx / sinx) and then i changed sec x to 1/ cosx so that i had ((1/cosx)/ sinx)  (cos x / sinx) after that i get stuck

calc
Where do I start to prove this identity: sinx/cosx= 1cos2x/sin2x please help!! Hint: Fractions are evil. Get rid of them. Well, cos2x = cos2x  sin2x, so 1coscx = 1  cos2x  sin2x = 1  cos2x + sin2x You should be able to

tigonometry
expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by b and using that cos(b)= cos(b) sin(b)= sin(b) gives: sin(ab) = sin(a)cos(b)  cos(a)sin(b)

Trig. Law of Cosines
Show that any triangle with standard labeling... a^2+b^2+c^2/2abc = cos(alpha)/a + cos(beta)/b + cos(gamma)/c I don't get it. Can someone please help me. Start here with the law of cosines: a^2 = b^2 + c^2 2bc Cos A b^2 = a^2 +

Precalculus
solve 4 sin^2x + 4 sqrt 2 cos x6=0 for all real values of x. Going by the example in the book I got to: 4 cos^2x + 4sqrt2 cosx 10 = 0 but do not know how to proceed. Any help would be great. Thanks

Math 12
Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer cscxtanx/sinx+cosx? Simplify #2: sin2x/1+cos2X = ??? I'm stuck on this one. I don't know what I should do.

Trigonometry Check
Simplify #3: [cosxsin(90x)sinx]/[cosxcos(180x)tanx] = [cosx(sin90cosxcos90sinx)sinx]/[cosx(cos180cosx+sinx180sinx)tanx] = [cosx((1)cosx(0)sinx)sinx]/[cosx((1)cosx+(0)sinx)tanx] = [cosxcosxsinx]/[cosx+cosxtanx] =
You can view more similar questions or ask a new question.