Dont copié formé thé book oh ya dont copié formé thé book cause you Oiil inde up being triste soi remembré not toi copié frome thé book oh ya!translation in english please!
0/5 points I Previous Answers My Compute the work done by the force F (sin(, y, in moving an object along the trajectory that is the line segment from (1, 1, ) to (2, 2, 2) followed by the line segment from (2,2, 2) to (-3, 6, 5) when force is measured in Newtons and distance in...
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 o...
2) Use the sum and difference identites sin[x + pi/4] + sin[x-pi/4] = -1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4 - cosx sin pi/4 = -1 2 sin x cos pi/4 =-1 cos pi/4 = sqr2/2 2sin^x(sqrt2/2) = -1 sin x = -sqrt2 x = 7pi/4 and 5pi/4 Am I correct?
prove these identiessin^2+tan^2=sec^2-cos^2 sin^2 sec^2 +sin^2=tan^2+sin^2
Helle, i need help for simplify the expression: [( cos x) ( sin x) - ( sin x) ( - sin x)] / ( sin x) ²
the original problem was: (sin x + cos x)^2 + (sin x - cos x)^2 = 2steps too pleaseI got 1 for (sin x + cos x)^2 but then what does (sin x - cos x)^2 become since it's minus?
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 identity of: sin^4x+cos^4x= (sin^2x+cos^2x)(sin^2x+cos^2x) sin^2x+cos^2=1 This is the answer I come up with, it is not one of the options available as an answer. The answers given are 1. -2sin^2xcos^2x 2. 1+2sin^2x-2sin^4x 3. 1+3sin^3x-2sin^2x 4. 1-2sin^2x+2sin^4x 5. 0
simplify sin 80 degrees- sin 10 degrees divided by sin 80 degrees+ sin 10 degrees , so that it involves a function of only one angle.
Pove that 1+sin(theta)divided by 1-Sin(theta)- 1-Sin(theta) Divided by 1+Sin(theta)= 4Tan(theta)times Sec(theta) Show step by step how to change the left side of the equation to equal the right side of the equation.