Rewrite #tantheta# and #cottheta# as sines and cosines using #color(magenta)(tan theta = sintheta/costheta and cot theta = costheta/sintheta#.
#(sin2theta)/(cos2theta) = 2/(costheta/sintheta - sintheta/costheta)#
I would recommend you simplify the right hand side prior to expanding the left.
#(sin2theta)/(cos2theta) = 2/((cos^2theta - sin^2theta)/(costhetasintheta)#
#(sin2theta)/(cos2theta) = (2costhetasintheta)/(cos^2theta - sin^2theta)#
We know this is true because #sin2theta= 2sinthetacostheta# and #cos2theta# can be written as #cos^2theta - sin^2theta#.
Practice exercises:
*Use the following table of trig identities to help you answer the next questions *
- Prove the following trig identities:
a) #(sin^2theta + cos^2theta + cot^2theta)/(1 + tan^2theta) = cot^2theta#
b) #cos(x + y) + cos(x - y) = 2cosxcosy#
c) #csc(2alpha) - cot(2alpha) = tan alpha#
Solve the following equation for x in the interval #0 ≤ x ≤ 2pi#:
#cos(2x) = 2sin^2x#
Hopefully this helps, and good luck!
