F.2數學(三角比的關係)20分~~

1.
Find the value of cos28sin62+cos62sin28
cos28sin62 + cos62sin28
= cos28sin(90-28) + sin28cos(90-62)
= cos^2 28 + sin^2 28
= 1//

2.
Prove the following identity.
(cosθ+sinθ)^2-1=2cos^2θ√[(1/cosθ)+1][(1/cosθ)-1]
RHS
= 2cos^2 θ √[(1/cosθ)+1][(1/cosθ)-1]
= 2cos^2 θ √[(1/cos^2 θ) - 1]
= 2cos^2 θ √[(1-cos^2 θ)/cos^2 θ]
= 2cos^2 θ √(sin^2 θ /cos^2 θ)
= 2cos^2 θ (sinθ/cosθ)
= 2sinθcosθ
= 1 + 2sinθcosθ - 1
= cos^2 θ + 2sinθcosθ + sin^2 θ - 1
= (cosθ + sinθ)^2 - 1
= LHS//

3.
Simplify sin(90-θ)/tan(90-θ)
sin(90-θ)/tan(90-θ)
= cosθ / (1/tanθ)
= cosθ / [1/(sinθ/cosθ)]
= cosθ / (cosθ/sinθ)
= cosθ x sinθ/cosθ
= sinθ//
4.
Prove the following identity.
sin^2θ-sin^4θ+cos^4θ=cos^2θ
LHS
= sin^2 θ (1 - sin^2 θ) + cos^4 θ
= (sin^2 θ)(cos^2 θ) + cos^4 θ
= cos^2 θ(sin^2 θ +cos^2 θ)
= cos^2 θ
= RHS//

1. cos28sin62 + cos62sin28
= sin ( 28 + 62 )
= sin 90
= 1

2. R.H.S.
= 2 cos^2θ √ [ ( 1 / cosθ ) + 1 ] [ ( 1 / cosθ ) - 1 ]
= 2 cos^2θ √ [ ( 1 / cosθ ) ^ 2 ] ( 1 + cosθ ) ( 1 - cosθ ) ------- 兩項都抽1/cosθ出黎
= 2 cosθ √ ( 1 + cosθ ) ( 1 - cosθ )
= 2 cosθ √ (1 - cos^2θ )
= 2 cosθ √ sin^2θ
= 2 cosθ sinθ
= 2 cosθ sinθ + 1 - 1
= 2 cosθ sinθ + cos^2θ + sin^2θ - 1 -------- 將1變做cos^2θ+sin^2θ
= ( cos^2θ + 2 cosθ sinθ + sin^2θ ) - 1 -------- 調位
= ( cosθ + sinθ )^2 - 1
= L.H.S.

3. sin ( 90 - θ ) / tan ( 90 - θ )
= sin ( 90 - θ ) cos ( 90 - θ ) / sin ( 90 - θ ) ]
= cos ( 90 - θ )
= sinθ

4. sin^2θ - sin^4θ + cos^4θ
= sin^2θ - sin^4θ + ( cos^2θ )^2
= sin^2θ - sin^4θ + ( 1 - sin^2θ )^2
= sin^2θ - sin^4θ + 1 - 2sin^2θ + sin^4θ
= 1 - sin^2θ
= cos^2θ

希望幫到你!! ^^

1.
Find the value of cos28sin62+cos62sin28

cos28sin62+cos62sin28=sin(62+28)
=sin90
=1
2.
Prove the following identity.
(cosθ+sinθ)^2-1=2cos^2θ√[(1/cosθ)+1][(1/cosθ)-1]
L.H.S.= (cosθ+sinθ)2-1
=( cos2θ+ sin2θ+2sinθcosθ)-1
=(1 +2sinθcosθ)-1
=2sinθcosθ
=2cos2θ(sinθ/cosθ)
=2cos2θ√[(sin2θ)/cos2θ]
=2cos2θ√[(1-cos2θ)/cos2θ]
=2cos2θ√[(1/cosθ)2-1]
2cos2θ√[(1/cosθ)+1][(1/cosθ)-1]
=R.H.S.
3.
Simplify sin(90-θ)/tan(90-θ)

sin(90-θ)/tan(90-θ)

=cosθ/cotθ
=sinθ
4.
Prove the following identity.
sin2θ-sin4θ+cos4θ=cos2θ
L.H.S.=sin2θ-sin4θ+cos4θ
=sin2θ-(sin4θ-cos4θ)
=sin2θ-(sin2θ-cos2θ)(sin2θ+cos2θ)
=sin2θ-(sin2θ-cos2θ)(1)
=cos2θ
=R.H.S.