F.3 Maths sin .cos. tan

(A) prove the following identities:
1. 1 - 2sin^2(x) = 2cos^2(x) - 1

L.H.S.
= 1 - 2sin^2(x)
= 1 - 2[1 - cos^2(x)]
= 1 - 2 + 2cos^2(x)
= 2cos^2(x) - 1
= R.H.S.


2. cos^2(x) - sin^2(x) = 2cos^2(x) - 1

L.H.S.
= cos^2(x) - sin^2(x)
= cos^2(x) - [1 - cos^2(x)]
= cos^2(x) - 1 + cos^2(x)
= 2cos^2(x) - 1
= R.H.S.


(B) Using the identities proved in (A) , show that
cos^4(x) - sin^4(x) = 1 - 2sin^2(x)

L.H.S.
= cos^4(x) - sin^4(x)
= [cos^2(x)]2 - [sin^2(x)]2
= [cos^2(x) + cos^2(x)][cos^2(x) - sin^2(x)]
= [1][cos^2(x) - sin^2(x)]
= cos^2(x) - sin^2(x)
= 2cos^2(x) - 1 .... [Refer to (A)2.]
= 1 - 2sin^2(x) .... [Refer to (A)1.]
= R.H.S.


(C) WITHOUT USING THE CALCULATOR AND SHOW YOUR STEPS , FIND THE VALUES OF THE FOLLOWING:
1) Incomplete question: sin^2(?)/cos60 - tan^2(60)/cos30


2) (1/tan60 + tan30)^2

(1/tan60 + tan30)^2
= [(1/√3) + (1/√3)]^2
= (2/√3)^2
= 4/3