the original problem was: (sin x + cos x)^2 + (sin x - cos x)^2 = 2 steps too please I got 1 for (sin x + cos x)^2 but then what does (sin x - cos x)^2 become since it's minus
for (sec x -1)(sec x + 1) = tan^(2) x so far I got up to: (sin^(2)x / cos x) (-sin^(2)x / cos x) what would the next step be? steps too please
the original problem was:Solve: sin(3x)-sin(x)=cos(2x)so far i've gooten to:sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x)Where would I go from here?
the original problem was: Solve: sin(3x)-sin(x)=cos(2x) so far i've gotten to: sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x) Where would I go from here?
I come to the answer #(-cos x sin^2 x)/((cos^2x -1)^2)# But it should be : #(-cos x )/((cos^2x -1))# I know #sin^2x = 1 - cos^2x# But its not the same as #cos^2x -1# So I do something wrong. My first step was : #((-2 cos x sinx sinx ) - (cos x (cos^2x -1)))/ (cos^2x -1)^2# Simplify #((-2cosx sin^2x) - cos x (cos^2x-1))/((cos^2x -1)^2)# Gives: #(-cosx((cos^2x-1)+2 sin^2x))/((cos^2x -1)^2)# Gives: #(-cosx(1+sin^2x-1))/((cos^2x -1)^2)# Gives: #(-cosxsin^2x)/((cos^2x -1)^2)#
Solve:(√2/2)/(-√2/2) steps tooand i got -1 but the answer is supposed to be 1
i was wondering if someone could check the answer i got for this math problem 7+2(5-2*9) = 243 I got -19I think you're supposed to do the parentheses first...but within that, you're supposed to do the multiplication and THEN the subtraction, and then mult it by 2....then add the 7....someone could check me on that too..it's been a while since i took algebra7+2(5-2*9) 7+2(5-18) 7+2(-13) 7-26 -197+2(5-2*9) 7+2(5-18) 7+2(-13) 7-26 -19
f(x) = x² + 2Cos²x, find f ' (x)a) 2(x+cos x)b) x - sin xc) 2x + sin xd)2(x - sin2x)I got neither of these answers, since the 2nd part should be chain rule, right?f(x) = x² + 2Cos²x = x² + 2(Cos x)²then f '(x) = (2)(2)(cosx)(-sinx)My answer:f '(x) = 2x - 4cosxsinxIf I do it this way, then I get that the answer is d:f(x) = x² + 2Cos²xf '(x) = x² + 2(-sin x)²f '(x) = 2x...
cos(x)=2 and sin(x)=5 Thanks for your previous answers mathmate. No. It's not a typo. I was wondering if someone could explain to me when in real life would I need to solve such problems, or similar, besides answering problems from a textbook. When ever I ask this question I never get a direct answer. Yes, trying to think of a real life example in which I would need to solve these type of problems does make my head hurt. I...
find all solutions in the interval [0,2 pi) sin(x+(3.14/3) + sin(x- 3.14/3) =1 sin^4 x cos^2 xSince sin (a+b) = sina cosb + cosb sina and sin (a-b) = sina cosb - cosb sina, the first problem can be written 2 sin x cos (pi/3)= sin x The solution to sin x = 1 is x = pi/2 For your other problem sin^4 cos^2 x = sin^4 x(1 - sin^2x)=0 The solutions are sin x = 0 and sin^2 x...
Solve this equation algebraically:(1-sin x)/cos x = cos x/(1+sin x)---I know the answer is an identity, and when graphed, it looks like cot x. I just don't know how to get there. I tried multiplying each side by its conjugate, but I still feel stuck. This is what I have so far:cos^2(x)/cos x + sin x = cos x - sin x/cos^2(x)...but I'm not really sure how to get to the answer. Help please?Thank you!