Group a
1. It is known that sin α = 55, sin (α-β) =- 10 10, and both α and β are acute angles, then β is equal to _ _ _ _ _.
Analysis: ∵ α and β are acute angles, ∴-π 2.
∵sinα=55,∴cosα= 1-(55)2 = 255。
∴sinβ=sin[α-(α-β)]=sinαcos(α-β)-cosαsin(α-β)=22.
∫0 & lt； β& lt； π2, ∴ β = π4. Answer: π 4
2. Known 0
Analysis: ∫0
∴cosβ= cos[(α+β)-α]= cos(α+β)cosα+sin(α+β)sinα=(-45)×35+(-35)×45 =-2425。 Answer: -2425.
3. If tanα and tanβ are two of the equations x2-3x-3 = 0, then sin (α+β) cos (α-β) = _ _ _ _ _.
Analysis: tan α+tan β = 3, tan α tan β =-3, then sin (α+β) cos (α-β) = sin α cos β+cos α sin β cos α cos β+sin α sin β.
= tanα+tanβ 1+tanαtanβ= 3 1-3 =-32。 Answer: -32.
4.Cos (α-π 6)+sin α = 453, then the value of sin (α+7π 6) is _ _.
Analysis: From the known 32cosα+ 12sinα+sinα = 453, that is, 12cosα+32sinα = 45,
Sin (α+π 6) = 45，SIN (α+76 π) =-SIN (α+π 6) =-45。 Answer: -45.
5. (Original question) Define operation a? B = A2-AB-B2, so sinπ 12? cosπ 12=________。
Analysis: sinπ 12? cosπ 12 = sin 2π 12-sinπ 12-cos 2π 12 =-(cos 2π 12-sin 2π 12)- 12×2。
6. It is known that α∈(π2, π), sin α 2+cos α 2 = 62.
(1) Find the value of cosα; (2) If sin (α-β) =-35, β∈(π2, π), find the value of cosβ.
Solution: (1) Because sin α 2+cos α 2 = 62, when both sides are squared at the same time, sin α = 12.
π 2
(2) Because π 2
Sin (α-β) =-35，COS (α-β) = 45。
cosβ= cos[α-(α-β)]= cosαcos(α-β)+sinαsin(α-β)
=-32×45+ 12×(-35)=-43+3 10.
Group b
1.cos2α 1+sin2α？ The value of 1+tan α 1-tan α is _ _ _ _ _.
Analysis: COS2α 1+SIN2α? 1+tanα 1-tanα= cos 2α-sin 2α(sinα+cosα)2？ 1+tanα 1-tanα
=cosα-sinαsinα+cosα？ 1+tanα 1-tanα= 1-tanα 1+tanα？ 1+tanα 1-tanα= 1。
2.Cos (π 4+x) = 35, then the value of sin2x-2sin2x 1-tanx is _ _ _ _ _.
Analysis: ∫cos(π4+x)= 35, ∴ cosx-sinx = 352.
∴ 1-sin2x= 1825,sin2x=725,∴sin2x-2sin2x 1-tanx=2sinx(cosx-sinx)cosx-sinxcosx=sin2x=725.
3. It is known that cos (α+π 3) = sin (α-π 3), then tan α = _ _ _ _ _ _
Analysis: cos (α+π 3) = cos α cos π 3-sin α sin π 3 =12 cos α-32 sin α, sin (α-π 3).
= sinαcosπ3-cosαsinπ3 = 12 sinα-32 cosα，
From: (12+32) sin α = (12+32) cos α, tan α = 1
4. Let α ∈ (π4,3π4), β ∈ (0,π 4), COS (α-π 4) = 35, SIN (3 π 4+β) = 5 13, then SIN (α+β) = _ _ _.
Analysis: α ∈ (π 4,3 π 4), α-π 4 ∈ (0,π 2), while cos (α-π 4) = 35, ∴ sin (α-π 4) = 45.
∵β∈(0,π4),∴3π4+β∈(3π4,π).∵sin(3π4+β)=5 13,∴cos(3π4+β)=- 12 13，
∴sin(α+β)=-cos[(α-π4)+(3π4+β)]
=-cos(α-π4)？ cos(3π4+β)+sin(α-π4)？ sin(3π4+β)=-35×(- 12 13)+45×5 13 = 5665，
That is, sin (α+β) = 5665.
5. It is known that cos α = 13, cos (α+β) =- 13, α, β∈(0, π2), then the value of cos (α-β) is equal to _ _ _ _ _.
Analysis: √α∈(0, π2), ∴2α∈(0, π). ∫cosα= 13，∴ Cos2α = 2cos2α- 1 =-79。 ∴cos(α-β)=cos[2α-(α+β)]=cos2αcos(α+β)+sin2αsin(α+β)=(-79)×(- 13)+429×223=2327.
6. Given the angle α in the first quadrant and cosα= 35°, then1+2cos (2α-π 4) sin (α+π 2) = _ _ _ _ _.
Analysis: ∵ α is in the first quadrant, and COS α = 35 and ∴ SIN α = 45, then1+2cos (2α-π 4) sin (α+π 2) =1+2 (22cos2alpha+22sin2α) cos.
7. It is known that a = (cos2α, sinα), b = (1, 2sinα- 1), α∈(π2, π), if a? B = 25, then the value of tan (α+π 4) is _ _ _ _ _.
Analysis: a? B = cos2α+2sin2α-sinα =1-2sin2α+2sin2α-sinα =1-sinα = 25, ∴ sinα = 35, and α∈(π2, π), ∴ cosα.
8. The value of tan10tan70tan70-tan10+tan120 is _ _ _ _.
Analysis: tan (70-10) = tan 70-tan101+tan 70? tan 10 =3，
Therefore, tan 70-tan10 = 3 (1+tan 70tan10) is substituted into the algebraic expression:
tan 70 tan 10 3( 1+tan 70 tan 10)+tan 120 = tan 70 tan 10 3( 1+tan 70 tan 10)-3 = tan 70 tan 10 3 tan 70 tan 10 = 33。
9. When the terminal edge of a given angle α passes through point A (-1, 15), the value of sin (α+π 4) sin2alpha+cos2alpha+1is equal to _ _ _ _ _.
Analysis: ∫sinα+cosα≠0, cos α =- 14, ∴ sin (α+π 4) sin 2 α+cos 2 α+1= 24 cos α =-2.
10. evaluation: cos20 sin20? cos 10+3 sin 10 tan 70-2cos 40。
Solution: the original formula = cos 20 cos10sin20+3sin10sin70cos70-2cos40.
= cos 20 cos 10+3 sin 10 cos 20 sin 20-2cos 40
= cos 20(cos 10+3 sin 10)sin 20-2cos 40
= 2cos 20(cos 10 sin 30+sin 10 cos 30)sin 20-2cos 40
= 2 cos 20 sin 40-2 sin 20 cos 40 sin 20 = 2。
1 1. Given the directional quantity m = (2cosx2, 1), n = (sinx2, 1) (x ∈ r), let the function f (x) = m? n- 1。
(1) Find the range of function f(x); (2) The three internal angles of acute angle △ABC are called A, B and C respectively. If f (a) = 5 13 and f (b) = 35, find the value of f(C).
Solution: (1) f (x) = m? n- 1=(2cosx2， 1)？ (sinx2， 1)- 1 = 2 cos x2 sin x2+ 1- 1 = sinx。
∵x∈R, ∴ The range of the function f(x) is [- 1, 1].
(2)∵f(a)=5 13,f(b)=35,∴sina=5 13,sinb=35.
∵A and B are acute angles, ∴ COSA =1-sin2a =1213, COSB = 1-SIN2B = 45.
∴f(c)=sinc=sin[π-(a+b)]=sin(a+b)=sinacosb+cosasinb
= 513× 45+1213× 35 = 5665. The value of ∴ f (c) is 5665.
12. Known: 0
(1) Find the value of sin2β; (2) Find the value of COS (α+π 4).
Solution: (1) Method 1: ∫ cos (β-π 4) = cos π 4 cos β+sin π 4 s in β = 22 cos β+22 sin β =13,
∴cosβ+sinβ=23,∴ 1+sin2β=29,∴sin2β=-79.
Method 2: sin2β = cos (π 2-2β) = 2cos2 (β-π 4)-1=-79.
(2)∵0 & lt； α& lt； π2 & lt； β& lt； π,∴π4<； β-π4 & lt； 3π4，π2 & lt； α+β& lt； 3π2,∴sin(β-π4)>； 0，cos(α+β)& lt； 0.
∵cos(β-π4)= 13,sin(α+β)=45,∴sin(β-π4)=223,cos(α+β)=-35.
∴cos(α+π4)=cos[(α+β)-(β-π4)]=cos(α+β)cos(β-π4)+sin(α+β)sin(β-π4)
=-35× 13+45×223=82-3 15.
The sum and difference of trigonometric functions in the second quarter: two angles and double angles
Group a
1. If sin α = 35, α ∈ (-π2, π2), then COS (α+5π 4) = _ _ _ _ _ _.
Analysis: Because α ∈ (-π2, π2) and sin α = 35, cos α = 45, it is obtained from the cosine formula of sum and difference of two angles: cos (α+5 π 4) =-22 (cos α-sin α) =-210.
2. Known π