SoftMath Algebrator - Algebra Homework Solver (689076614429) "sin" Questions

Trig Identities cosecӨ-sinӨ=cos²Ө/sinӨ Change csc to 1/sin. Find a common denominator and add the two left terms. 1/sin - sin = (1 -sin^2)/sin. Rewrite formula (1 - sin^2)/sin

Differentiate each of the following w.r.t.x; 29.sin2xsinx Use the rule for differentiating products of functions: ()' signifies derivative (29*sin(2X)*sin(X))'= (29)'*sin(2X)*sin(X) +29* ...

1 + sinx = 2cos^2x Hii..this can be done as follow cos^2 x = 1 - sin^2 x--------(1) 2cos^2 x=1+ sin x 2cos^2 x - sin x -1=0 Substituting formula (1) 2(1 - sin^2 x) - sin x - 1 = 0 2sin^2 x + sin x - 1

...= 1/cot^4X + 1/cot^2X sec^4X- sec^2X = 1/cot^4X + 1/cot^2X RHS 1/cot^4X + 1/cot^2X =1/(Cos^4X/Sin^4X) + 1/(Cos^2X/Sin^2X) =Sin^4X/Cos^4X + Sin^2X/Cos^2X =Sin^4X/Cos^4X + Cos^2X.Sin^2X/Cos^4X ...

(sin-cos)(sin+cos) help!! (sinX-cosX)(sinX+cosX) =(sin^2 - Cos^2X) =Sin^2X - (1-Sin^2X) =Sin^2X -1 + Sin^2X =2Sin^2X -1 Zulfikar Ali [email protected] 9899780221

why limit of sin x is 1, when x tends to 0 ? Sin 0 = 0 Sin 90 = 1 Sin 180 = 0 Sin 270 = -1 Sin 360 = Sin 0 = 0 Is there an angle greater than 360??? Is there any greater value of Sine than 1 ???

find the trig function given its period cos 5 pi also need the answer for sin 19pi/6 cos(5PI)=cos(4PI+PI)=cos(PI)=-1sin(19PI/6)=sin(18PI/6+ PI/6)=sin(3PI +Pi/6)=sin(2PI+PI+PI/6)=sin(PI+PI/6)=sin(-PI/6

solve sin(A+B) if sin A=1/2, and sin B=1/2 From the given data we can get angle A=30 degrees and B = 30 degrees. There fore sin ( A+B) = sin ( 30+30) =sin 60 = root ( 3 )/ 2

Solve it plz...... cos x + root 3 sin x =root 2 cos x + root 3 sin x =root 2 cos x + ?3 * Sin x = ?2 squaring both the side (cos x + ?3 * Sin x)2 = (?2)2 Cos2 x + 3 * Sin2 x = 2 Cos2 x + Sin2 x + 2 * ...

sin x cos x = -1/2 solve for x sin x cos x = -1/2 solve for x sin x cos x = -1/2 => 2sinx cosx = -1 => sin(2x) = -1 => 2x = (3pi)/2 OR 2x = 270° => x = 3pi/4 OR x = 135°