Questions & Answers for: cos a sin a

sin, cos, etc. Is there a way to do this....... if you want to square sin(3), press: ( sin 3) )^ 2 1) Selecting the sin/cos/tan key results in sin( or cos( or tan( being displayed on your computer, 2)

...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 = cos^2/sin Divide out the ...

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 * Sin2 x = 2 1 + 2 * Sin2 x= 2 2 * Sin2 x = 2-1 2 * Sin2 x = 1 Sin2 x = ½ Sin x

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

cos+tan(sin)=sec This is a trigonometry problem not a calculator's. cos(x) +tan(x).sin(x) = ( cos(x) + (sin(x)/cos(x)).sin(x). After reduction to the same denominator (which is cos(x)) you obtain {cos

sin(x))= cos(x) d^2/dx^2 (sin(x))= d/dx( cos(x))=-sin(x) d^3/dx^3 (sin(x))= -cos(x) d^4/dx^4 (sin(x))=d/dx(-cos(x))=sinx. In 100 successive differentiations you perform the foregoing sequence 100/4=25

sin(2X))'=cos(2X)*(2X)'=2*cos(2X) (sin(X))'=cos(X)Result is (29*sin(2X)*sin(X))'= 29*2*cos(2X)*sin(X)+29*sin(2X)*cos(X) You could also have cast your formula in the form sin(2X)*sin(X)= 1/2[ cos(2X-X)

I need to rewrite Y=5(sqrt2)sin(x)-5(sqrt2)cos(x) as Y=Asin(Bx-c) Use the fact that cos(pi/4)=sin(pi/4)= 1/square root(2). Trigonometric identity cos(a+b)=cos(a)cos(b)-sin(a)sin(b).

...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 ...

...sin(a))/cos(a) ln(sec(a)+tan(a))= ln( (1+sin(a))/cos(a))=X 2*cosh(X)= e^(X)+e^(-X) e^(X)=(1+sin(a))/cos(a) e^(-X)= cos(a)/(1+sin(a)) 2cosh(X)=(1+sin(a))/cos(a) +cos(a)/(1+sin(a))= 2/cos(a) ...