i have some problems doing trig the first one is: Show that cos(x/2) sin(3x/2) = ½(sinx + sin2x) i know that you are supposed to substitute all those trig function things in it but i kind of forgot how to the only that i can see
I can't seem to prove these trig identities and would really appreciate help: 1. cosx + 1/sin^3x = cscx/1 - cosx I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1-cosx Simplified: cosx + sin^3x/sin^3x = cscx/1-cosx I don't
Use a trig identity to combine two functions into one so you can solve for x. (The solution should be valid for any value of t). 3cos(t) + 3*sqrt(3)*sin(t)=6cos(t-x) I know that 6 cos(t-x) can be 6(cos(t)cosx(x)+sin(t)sin(x)) I
If secx = 8 and -pi/2 < x < 0, find the exact value of sin2x Use the identity sin 2x = 2(sinx)(cosx) if secx = 8, then cosx = 1/8 where x is in the fourth quadrant. consider a right angled triangle with x=1, r=8, then y=??
I need to prove that the following is true. Thanks (cosx / 1-sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds of problems when I was still teaching. it requires a little "trick" L.S. =cosx/(1-sinx) multiply top
prove the identity (sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2 sinX^6= sinx^2 ^3 = (1-cosX^2)^3 = (1-2CosX^2 + cos^4) (1-cosX^2) then multiply that out 1-2CosX^2 + cos^4 - cosX^2 + 2cos^4 -cos^6 add that on the left to the cos^6,