回答:
下記参照
説明:
Sin ^ 4x = 1/8(3-4cos2x + cos4x)証明するには?
LHS = sin ^ 4x = [(2sin ^ 2x)/ 2] ^ 2 = 1/4 [(1-cos2x)^ 2] = 1/4 [1-2cos2x + cos ^ 2(2x)] = 2 /( 4 * 2)[1-2cos2x + cos ^ 2(2x)] = 1/8 [2-4cos2x + 2cos ^ 2(2x)] = 1/8 [2-4cos2x + 1 + cos4x] = 1/8 [ 3-4cos2x + cos4x] = RHS
証明するには? TanA(1 + Sec2A)= Tan2A
LHS = tanA(1 + sec2A)= tanA(1 + 1 /(cos2A))= tanA((1 + cos2A)/(cos2A))= 1 /(cos2A)[tanA * 2cos ^ 2A] = 1 /(cos2A) )[sinA / cosA * 2cos ^ 2A] =(sin2A)/(cos2A)= RHS