I will use x instead of theta for easier typing
2 sin^2 x - 1 = cos 2x
do you have to prove this?
if so, then
RS = -cos(2x)
= -cos(x+x)
= -cosxcosx + sinxsinx
= -cos^x + sin^x
= -(1-sin^x_ + sin^x
= 2sin^x - 1
= LS
Please help...I'm not understanding trig identities and how to manipulate and express these two problems in their associated functions.
Thanks
a)Express as a function of cos (theta)
2 sin^2(theta) - 1
b)Express as a function of sin (theta) or cos (theta)
tan^2(theta) - 2 sec(theta)sin(theta)
2 answers
for the second
tan^x - 2secxsinx
= sin^2 x/cos^2 x - 2sinx/cosx
= sin^2 x/cos^2 x 2 sinxcosx/cos^2 x
= (sin^2 x - sin 2x)/(1 - sin^2 x)
this is in terms of sinx
tan^x - 2secxsinx
= sin^2 x/cos^2 x - 2sinx/cosx
= sin^2 x/cos^2 x 2 sinxcosx/cos^2 x
= (sin^2 x - sin 2x)/(1 - sin^2 x)
this is in terms of sinx